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      "rank": 1,
      "url": "https://www.potomacindexing.com/estimating-index-size/",
      "title": "Estimating Index Size - Potomac Indexing, LLC",
      "content": "Chicago Manual of Style is much simpler: <strong>5 entries per page = an index 1/50th the length of the text</strong>. 15+ entries = 1/20th the length of the text. index pages 1 \u2014\u2014\u2014\u2014\u2014\u2013 = \u2014\u2014 text pages x\nOne of the best ways to cut down on editing time is to make sure your index is the right size in the first place. Not all clients will give you a length requirement to fill, but when they do, it&#8\u2026\ntotal index lines \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2013 = entries per page (assuming one locator per entry) text pages \u00b7 Then, subtract 20% for turnover lines. You still will be able to have multiple locators per heading, so it usually comes out more or less in the wash. Chicago Manual of Style is much simpler: 5 entries per page = an index 1/50th the length of the text. 15+ entries = 1/20th the length of the text. index pages 1 \u2014\u2014\u2014\u2014\u2014\u2013 = \u2014\u2014 text pages x\nThis works great if you know the number of pages allotted for the index. If you\u2019re given the number of lines instead, you can either look up the table in Wellisch, or just go to another book by the same publisher and manually count index lines. Amazon\u2019s Look Inside feature or Google Books can save you a trip to the library.\nGo with Mulvaney\u2019s recommendation, and keep an eye on your number of entries per page. In my experience, scholarly introductions are likely to not be indicative of the rest of the book\u2019s density, so you may not get an accurate picture until you\u2019ve done the first couple of chapters and can get an average. The point is to be close enough to have something to work with when it comes down to editing. Changing an index length by 1 or 2% is wholly different from changing it by 50%, and can mean the difference between making or missing a deadline.",
      "content_chars": 1703,
      "published_date": "2022-10-04T19:26:03"
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      "rank": 2,
      "url": "https://learn.microsoft.com/en-us/sql/relational-databases/databases/estimate-the-size-of-a-nonclustered-index?view=sql-server-ver16",
      "title": "Estimate the Size of a Nonclustered Index - SQL Server | Microsoft Learn",
      "content": "Allocation pages: There&#x27;s at least one IAM page used to track the pages allocated to a heap, but the space overhead is minimal and there&#x27;s no algorithm to deterministically calculate exactly how many IAM pages to use. Large object (LOB) values: The algorithm to determine exactly how much space will be used to store the LOB data types varchar(max), varbinary(max), nvarchar(max), text, ntext, xml, and image values is complex. It&#x27;s sufficient to just add the average size of the LOB values expected, multiply by Num_Rows, and add that to the total nonclustered index size.\nThe 2 in the formula is for the row's entry in the page's slot array. Calculate the number of pages required to store all the rows: Num_Leaf_Pages = Num_Rows / (Leaf_Rows_Per_Page - Free_Rows_Per_Page) The number of pages estimated should be rounded up to the nearest whole page. Calculate the size of the index (8,192 total bytes per page):\nNum_Index_Pages = 1000/(25^3)+ 1000/(25^2) + 1000/(25^1) = 1 + 2 + 40 = 43, which is the number of pages described in the example. Calculate the size of the index (8,192 total bytes per page):\nUse this procedure to estimate the amount of space that is required to store a nonclustered index in SQL Server.\nAllocation pages: There's at least one IAM page used to track the pages allocated to a heap, but the space overhead is minimal and there's no algorithm to deterministically calculate exactly how many IAM pages to use. Large object (LOB) values: The algorithm to determine exactly how much space will be used to store the LOB data types varchar(max), varbinary(max), nvarchar(max), text, ntext, xml, and image values is complex. It's sufficient to just add the average size of the LOB values expected, multiply by Num_Rows, and add that to the total nonclustered index size.",
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      "published_date": "2024-11-22T00:00:00"
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      "rank": 3,
      "url": "https://learn.microsoft.com/en-us/sql/relational-databases/databases/estimate-the-size-of-a-clustered-index?view=sql-server-ver17",
      "title": "Estimate the Size of a Clustered Index - SQL Server | Microsoft Learn",
      "content": "Calculate the amount of space that is required to store the data in the leaf level <strong>(8,192 total bytes per page</strong>): ... You can use the following steps to estimate the amount of space that is required to store the upper levels of the index: Specify ...\nCalculate the amount of space that is required to store the data in the leaf level (8,192 total bytes per page): ... You can use the following steps to estimate the amount of space that is required to store the upper levels of the index: Specify the number of fixed-length and variable-length columns in the index key and calculate the space that is required for their storage: The key columns of an index can include fixed-length and variable-length columns. To estimate the interior level index row size, calculate the space that each of these groups of columns occupies within the index row.\nNum_Index_Pages = 1000/(25^3)+ 1000/(25^2) + 1000/(25^1) = 1 + 2 + 40 = 43, which is the number of pages described in the example. Calculate the size of the index (8,192 total bytes per page):\nUse this procedure to estimate the amount of space that is required to store data in a clustered index in SQL Server.\nThe space overhead is minimal, and there's no algorithm to deterministically calculate exactly how many IAM pages will be used. Large object (LOB) values: The algorithm to determine exactly how much space will be used to store the LOB data types varchar(max), varbinary(max), nvarchar(max), text, ntext, xml, and image values is complex. It's sufficient to just add the average size of the LOB values that are expected, multiply by Num_Rows, and add that to the total clustered index size.",
      "content_chars": 1662,
      "published_date": "2025-11-18T00:00:00"
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    {
      "rank": 4,
      "url": "https://www.olostep.com/",
      "title": "Web Data Infrastructure for AI Agents | Olostep",
      "content": "Turn any URL into LLM-ready Markdown, HTML, screenshots, PDFs, or structured JSON. Handle JS-rendered pages, actions, and extraction workflows without maintaining browsers, proxies, or brittle scrapers. Crawl websites at scale, collect content from subpages, control depth and URL patterns, and retrieve clean HTML or Markdown for indexing, enrichment, RAG, and AI workflows.\nTurn any URL into LLM-ready Markdown, HTML, screenshots, PDFs, or structured JSON. Handle JS-rendered pages, actions, and extraction workflows without maintaining browsers, proxies, or brittle scrapers. Crawl websites at scale, collect content from subpages, control depth and URL patterns, and retrieve clean HTML or Markdown for indexing, enrichment, RAG, and AI workflows.\nSee how Olostep powers AI platforms, sales lead enrichment, deep research, competitive intelligence, and SEO teams with one API. Access custom, hyper-specialized B2B indexes for your industry to search and extract comprehensive data beyond what general web indexes cover\nDiscover every URL on a website using sitemaps and on-page links. Filter by path patterns, paginate large results, and prepare clean URL lists for SEO, crawls, and batches. Process up to 10k concurrent URLs in a single batch in 5-8 mins to get clean web data and aggregate content. Run many batches in parallel to scale to millions of concurrent requests. Ask natural-language questions and get AI-generated answers grounded in web sources.\nCreate scheduled web monitors from natural-language instructions. Track for changes on a single page or across the web, deltas, extract structured insights, and get chance notifications by email, webhook, or SMS.",
      "content_chars": 1676,
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    },
    {
      "rank": 5,
      "url": "https://www.sqlskills.com/blogs/paul/on-index-key-size-index-depth-and-performance/",
      "title": "On index key size, index depth, and performance - Paul S. Randal",
      "content": "So with a index key size of 8 bytes, the index needs <strong>4214 pages</strong> at level 1 in the index structure to hold references to all 2 million leaf-level pages. This means the fanout value is 2000000 / 4214, which is approximately 474. The times for the 2 million selects for the 8-byte cluster key were ...\nFor the same data, if the index fanout is changed to 100, and the index has 10,000 pages at the leaf level, the next level needs 100 pages, and then there\u2019s the root page. That\u2019s a total of only three levels. I want to measure whether there\u2019s a noticeable performance difference based on the fanout, and hence index depth, of an index from varying it\u2019s key size, for single-row select operations. There won\u2019t be any noticeable effect on scans, as that only involves a single traversal of the index, to find the starting point of the scan.\nSo with a index key size of 8 bytes, the index needs 4214 pages at level 1 in the index structure to hold references to all 2 million leaf-level pages. This means the fanout value is 2000000 / 4214, which is approximately 474. The times for the 2 million selects for the 8-byte cluster key were 21.983s, 21.94s, 21.973s, 21.967s, 21.963s, with an average of 21.9652s, and a per-select average of 10.98 microseconds. Running the test for each of my test key sizes produced the following results: Key Size Index Depth Total Page Count Fanout Average Time for selects Rough time per select -------- ----\nNo \u2013 it can\u2019t. The index levels only say what\u2019s the lowest key value that can appear on the page below, not that any particular key exists at the leaf level. ... Would the use of included columns, instead of one or more key columns, mitigate this at all? Thanks. ... For nonclustered indexes, absolutely! As only the index key is pushed into the upper levels of the index, the smaller the total key length, the better. ... Key Size Index Depth Total Page Count Fanout Average Time for selects Rough time per select.\nIn my Insider newsletter a couple of weeks ago, I discussed how index fragmentation is often considered when designing indexes, but index depth often isn\u2019t. In the newsletter I said I\u2019d do a more comprehensive blog post with some data, so this is it. Fanout and Index Depth The index depth is determined by the fanout of [\u2026]",
      "content_chars": 2296,
      "published_date": "2015-10-26T13:47:41"
    },
    {
      "rank": 6,
      "url": "https://stackoverflow.com/questions/781873/how-to-figure-out-size-of-indexes-in-mysql",
      "title": "How to figure out size of Indexes in MySQL - Stack Overflow",
      "content": "If you are using InnoDB tables, you can get the size for individual indexes from mysql.innodb_index_stats. The &#x27;size&#x27; stat contains the answer, in pages, so you have to multiply it by the page-size, which is <strong>16K by default</strong>.\nThis gives the total index size, not a size per index (assuming a table has multiple indices). 2015-04-08T14:54:34.803Z+00:00 ... Save this answer. ... Show activity on this post. If you are using InnoDB tables, you can get the size for individual indexes from mysql.innodb_index_stats. The 'size' stat contains the answer, in pages, so you have to multiply it by the page-size, which is 16K by default.\nOn MyISAM, each index block is 4 KB page filled up to fill_factor with index records, each being key length + 4 bytes long. ... As for InnoDB, the table is always clustered on the PRIMARY KEY, there is no separate PRIMARY KEY index ... Save this answer. ... Show activity on this post. Using phpMyAdmin, when viewing the table structure there is a Details link at the bottom somewhere. Once you click on it it will show you the total size of the indexes you have on the table where it is marked Space Usage.\nin my case it gave a very different (and unrealistic) size from the one reported by show table status from [dbname] 2017-07-28T07:41:56.413Z+00:00 ... Save this answer. ... Show activity on this post. Here's an adaption from some of the above to also give you the percentage of the total index for the table that each index has used, hopefully this will be useful for someone \u00b7 select database_name, table_name, index_name, round((stat_value*@@innodb_page_size)/1024/1024, 2) SizeMB, round(((100/(SELECT INDEX_LENGTH FROM INFORMATION_SCHEMA.TABLES t WHERE t.TABLE_NAME = iis.table_name and t.TABLE_SCHEMA = iis.database_name))*(stat_value*@@innodb_page_size)), 2) `Percentage` from mysql.innodb_index_stats iis where stat_name='size' and table_name = 'TargetTable' and database_name = 'targetDB'\nThis is how you can get all of the indexes size, in megabytes, without the PRIMARY (which is the table itself), ordered by size. SELECT database_name, table_name, index_name, ROUND(stat_value * @@innodb_page_size / 1024 / 1024, 2) size_in_mb FROM mysql.innodb_index_stats WHERE stat_name = 'size' AND index_name != 'PRIMARY' ORDER BY size_in_mb DESC;",
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    },
    {
      "rank": 7,
      "url": "https://docs.olostep.com/get-started/welcome",
      "title": "Welcome to Olostep - Olostep Docs",
      "content": "Fetch the complete documentation index at: /llms.txt \u00b7 Use this file to discover all available pages before exploring further. ... Olostep: Infrastructure for the Web\u2019s second user.\nFetch the complete documentation index at: /llms.txt \u00b7 Use this file to discover all available pages before exploring further. ... Olostep: Infrastructure for the Web\u2019s second user.\nThe Olostep Agent allows to automate research workflows in a no code way with just a prompt in natural language. For AI agents: fetch docs.olostep.com/llms.txt for a complete index of this documentation before exploring further or get started here.\nDrop-in skills that teach AI coding agents how and when to use Olostep. Install with olostep add skills. Pull any URL as clean Markdown, HTML, screenshots, or structured JSON. Recursively gather every page on a site, with filters and search.\nLive web search with structured links and optional inline scraping. Self-healing extractors that turn pages into typed JSON at scale.",
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      "published_date": "2026-06-12T12:41:55"
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      "rank": 8,
      "url": "https://docs.opensearch.org/latest/api-reference/list/list-indices/",
      "title": "List Indices API | OpenSearch Documentation",
      "content": "{ &quot;next_token&quot;: &quot;MTcyOTE5NTQ5NjM5N3wub3BlbnNlYXJjaC1zYXAtbG9nLXR5cGVzLWNvbmZpZw==&quot;, &quot;indices&quot;: [ { &quot;health&quot;: &quot;green&quot;, &quot;status&quot;: &quot;open&quot;, &quot;index&quot;: &quot;movies&quot;, &quot;uuid&quot;: &quot;UZbpfERBQ1-3GSH2bnM3sg&quot;, &quot;pri&quot;: &quot;1&quot;, &quot;rep&quot;: &quot;1&quot;, &quot;docs.count&quot;: &quot;1&quot;, &quot;docs.deleted&quot;: &quot;0&quot;, &quot;store.size&quot;: &quot;7.7kb&quot;, &quot;pri.store.size&quot;: &quot;3.8kb&quot; } ] } ... Thank you for your feedback! Have a question? Ask us on the OpenSearch forum. Want to contribute? Edit this page or create an issue.\nThe index\u2019s health status. ... When using the next_token path parameter, use the token produced by the response to see the next page of indexes.\n{ \"next_token\": \"MTcyOTE5NTQ5NjM5N3wub3BlbnNlYXJjaC1zYXAtbG9nLXR5cGVzLWNvbmZpZw==\", \"indices\": [ { \"health\": \"green\", \"status\": \"open\", \"index\": \"movies\", \"uuid\": \"UZbpfERBQ1-3GSH2bnM3sg\", \"pri\": \"1\", \"rep\": \"1\", \"docs.count\": \"1\", \"docs.deleted\": \"0\", \"store.size\": \"7.7kb\", \"pri.store.size\": \"3.8kb\" } ] } ... Thank you for your feedback! Have a question? Ask us on the OpenSearch forum. Want to contribute? Edit this page or create an issue.\nThe amount of disk space used by the index.\nThe number of shards contained in the index.",
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      "rank": 9,
      "url": "https://medium.com/@davidfagb/turning-the-web-into-a-real-time-database-with-olostep-f175e956a603",
      "title": "Turning the Web Into a Real-Time Database with OloStep | by David Fagbuyiro | Medium",
      "content": "Instead of scraping, crawling, or building static indexes, OloStep flips the model. It treats the open web itself as a live, structured, and queryable database in real time. ... The internet is one of the most extensive and valuable datasets in history. However, it was not designed for structure, but it was created for publishing. There are billions of pages, including blogs, product listings, pricing tables, and job boards, which are constantly being updated.\nYou\u2019re building an AI assistant for a startup CEO who needs to monitor competitors, track pricing updates, new hires, and product launches\u2026\nInstead of scraping, crawling, or building static indexes, OloStep flips the model. It treats the open web itself as a live, structured, and queryable database in real time. ... The internet is one of the most extensive and valuable datasets in history. However, it was not designed for structure, but it was created for publishing. There are billions of pages, including blogs, product listings, pricing tables, and job boards, which are constantly being updated.\nOlostep is a powerful web scraping API that efficiently provides e data from any website. It meets the essential demands for fast, reliable, and cost-effective data acquisition, AI development and large-scale data aggregation. rtups, and established companies enhance their data-driven applications and automate workflows. Think SQL, but the tables are web pages.\nSide-by-side comparison \u2014 Left: messy HTML page, Right: clean JSON-like table extracted live]. Most tools that extract data from the web fall into one of two buckets: Scrapers: Fast but fragile. Break often, need constant maintenance. Knowledge graphs: Structured but stale. Require time-consuming ingestion pipelines. OloStep is different because it utilizes AI-native tools to extract structured data pages in real-time.",
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      "published_date": "2025-10-08T13:29:32"
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      "url": "https://learn.microsoft.com/en-us/sql/relational-databases/pages-and-extents-architecture-guide?view=sql-server-ver17",
      "title": "Page and Extent Architecture Guide - SQL Server | Microsoft Learn",
      "content": "Similar to a book, the Database Engine writes all data rows on pages. The size of every page is the same: <strong>8 KiB</strong>. In a book, most pages contain the data, or the main content of the book.\nOtherwise, the row contains a 16-byte pointer to a separate tree of text/LOB pages storing the LOB data in a LOB_DATA allocation unit. The large value types out of row table option controls this behavior. The 8,060-byte restriction is relaxed for tables and indexes that contain variable length columns using the varchar, nvarchar, varbinary, sql_variant, or CLR user-defined data types. When the total row size of all fixed and variable length columns in a heap or index exceeds the 8,060-byte limitation, the Database Engine dynamically moves one or more variable length columns to pages in a ROW_OVERFLOW_DATA allocation unit, starting with the widest column.\nSimilar to a book, the Database Engine writes all data rows on pages. The size of every page is the same: 8 KiB. In a book, most pages contain the data, or the main content of the book. Some pages contain metadata describing the content, for example, the table of contents and the index.\nAs a result, rows on a page are also accessed in a logical order, for example the order defined by the key of the BTree index that owns the page. To support large rows that don't fit on a single page, the part of the row that doesn't fit can be stored on other pages. The maximum size of data and overhead that can be contained in a single row on a page is 8,060 bytes.\nAn index or partition has row-overflow data if the function returns rows where the alloc_unit_type_desc column is ROW_OVERFLOW_DATA and the page_count column is greater than 0. An extent is a collection of eight physically contiguous pages. The size of each extent is 64 KiB.",
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              "text": "<p>Extending Vajk Hermecz's answer.<br>\nThis is how you can get all of the indexes size, in megabytes, without the PRIMARY (which is the table itself), ordered by size.</p>\n\n<pre><code>SELECT database_name, table_name, index_name,\nROUND(stat_value * @@innodb_page_size / 1024 / 1024, 2) size_in_mb\nFROM mysql.innodb_index_stats\nWHERE stat_name = 'size' AND index_name != 'PRIMARY'\nORDER BY size_in_mb DESC;\n</code></pre>",
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            "One of the best ways to cut down on editing time is to make sure your index is the right size in the first place. Not all clients will give you a length requirement to fill, but when they do, it&#8\u2026",
            "total index lines \u2014\u2014\u2014\u2014\u2014\u2014\u2014\u2013 = entries per page (assuming one locator per entry) text pages \u00b7 Then, subtract 20% for turnover lines. You still will be able to have multiple locators per heading, so it usually comes out more or less in the wash. Chicago Manual of Style is much simpler: 5 entries per page = an index 1/50th the length of the text. 15+ entries = 1/20th the length of the text. index pages 1 \u2014\u2014\u2014\u2014\u2014\u2013 = \u2014\u2014 text pages x",
            "This works great if you know the number of pages allotted for the index. If you\u2019re given the number of lines instead, you can either look up the table in Wellisch, or just go to another book by the same publisher and manually count index lines. Amazon\u2019s Look Inside feature or Google Books can save you a trip to the library.",
            "Go with Mulvaney\u2019s recommendation, and keep an eye on your number of entries per page. In my experience, scholarly introductions are likely to not be indicative of the rest of the book\u2019s density, so you may not get an accurate picture until you\u2019ve done the first couple of chapters and can get an average. The point is to be close enough to have something to work with when it comes down to editing. Changing an index length by 1 or 2% is wholly different from changing it by 50%, and can mean the difference between making or missing a deadline."
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            "The 2 in the formula is for the row's entry in the page's slot array. Calculate the number of pages required to store all the rows: Num_Leaf_Pages = Num_Rows / (Leaf_Rows_Per_Page - Free_Rows_Per_Page) The number of pages estimated should be rounded up to the nearest whole page. Calculate the size of the index (8,192 total bytes per page):",
            "Num_Index_Pages = 1000/(25^3)+ 1000/(25^2) + 1000/(25^1) = 1 + 2 + 40 = 43, which is the number of pages described in the example. Calculate the size of the index (8,192 total bytes per page):",
            "Use this procedure to estimate the amount of space that is required to store a nonclustered index in SQL Server.",
            "Allocation pages: There's at least one IAM page used to track the pages allocated to a heap, but the space overhead is minimal and there's no algorithm to deterministically calculate exactly how many IAM pages to use. Large object (LOB) values: The algorithm to determine exactly how much space will be used to store the LOB data types varchar(max), varbinary(max), nvarchar(max), text, ntext, xml, and image values is complex. It's sufficient to just add the average size of the LOB values expected, multiply by Num_Rows, and add that to the total nonclustered index size."
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            "Calculate the amount of space that is required to store the data in the leaf level (8,192 total bytes per page): ... You can use the following steps to estimate the amount of space that is required to store the upper levels of the index: Specify the number of fixed-length and variable-length columns in the index key and calculate the space that is required for their storage: The key columns of an index can include fixed-length and variable-length columns. To estimate the interior level index row size, calculate the space that each of these groups of columns occupies within the index row.",
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            "Use this procedure to estimate the amount of space that is required to store data in a clustered index in SQL Server.",
            "The space overhead is minimal, and there's no algorithm to deterministically calculate exactly how many IAM pages will be used. Large object (LOB) values: The algorithm to determine exactly how much space will be used to store the LOB data types varchar(max), varbinary(max), nvarchar(max), text, ntext, xml, and image values is complex. It's sufficient to just add the average size of the LOB values that are expected, multiply by Num_Rows, and add that to the total clustered index size."
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                "question": "What is counted as a request?",
                "answer": "One request equals one webpage or one PDF processed. We do not charge separately for bandwidth, proxies, or data usage. All infrastructure costs are included in the price per request.",
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                "answer": "No, Olostep does not charge for failed requests. You are only billed for successful requests, ensuring predictable and fair usage-based pricing. For endpoints that involve LLM processing (such as the Answers API), any underlying model costs may still apply. However, Olostep itself only charges for requests that are successfully completed.",
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                "question": "Is web scraping legal?",
                "answer": "Web scraping is legal in many cases, but depends on how the data is accessed and used. It is important to follow website terms of service, data privacy regulations and applicable laws when extracting web data.",
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            "Turn any URL into LLM-ready Markdown, HTML, screenshots, PDFs, or structured JSON. Handle JS-rendered pages, actions, and extraction workflows without maintaining browsers, proxies, or brittle scrapers. Crawl websites at scale, collect content from subpages, control depth and URL patterns, and retrieve clean HTML or Markdown for indexing, enrichment, RAG, and AI workflows.",
            "See how Olostep powers AI platforms, sales lead enrichment, deep research, competitive intelligence, and SEO teams with one API. Access custom, hyper-specialized B2B indexes for your industry to search and extract comprehensive data beyond what general web indexes cover",
            "Discover every URL on a website using sitemaps and on-page links. Filter by path patterns, paginate large results, and prepare clean URL lists for SEO, crawls, and batches. Process up to 10k concurrent URLs in a single batch in 5-8 mins to get clean web data and aggregate content. Run many batches in parallel to scale to millions of concurrent requests. Ask natural-language questions and get AI-generated answers grounded in web sources.",
            "Create scheduled web monitors from natural-language instructions. Track for changes on a single page or across the web, deltas, extract structured insights, and get chance notifications by email, webhook, or SMS."
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        {
          "title": "On index key size, index depth, and performance - Paul S. Randal",
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          "description": "So with a index key size of 8 bytes, the index needs <strong>4214 pages</strong> at level 1 in the index structure to hold references to all 2 million leaf-level pages. This means the fanout value is 2000000 / 4214, which is approximately 474. The times for the 2 million selects for the 8-byte cluster key were ...",
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            "For the same data, if the index fanout is changed to 100, and the index has 10,000 pages at the leaf level, the next level needs 100 pages, and then there\u2019s the root page. That\u2019s a total of only three levels. I want to measure whether there\u2019s a noticeable performance difference based on the fanout, and hence index depth, of an index from varying it\u2019s key size, for single-row select operations. There won\u2019t be any noticeable effect on scans, as that only involves a single traversal of the index, to find the starting point of the scan.",
            "So with a index key size of 8 bytes, the index needs 4214 pages at level 1 in the index structure to hold references to all 2 million leaf-level pages. This means the fanout value is 2000000 / 4214, which is approximately 474. The times for the 2 million selects for the 8-byte cluster key were 21.983s, 21.94s, 21.973s, 21.967s, 21.963s, with an average of 21.9652s, and a per-select average of 10.98 microseconds. Running the test for each of my test key sizes produced the following results: Key Size Index Depth Total Page Count Fanout Average Time for selects Rough time per select -------- ----",
            "No \u2013 it can\u2019t. The index levels only say what\u2019s the lowest key value that can appear on the page below, not that any particular key exists at the leaf level. ... Would the use of included columns, instead of one or more key columns, mitigate this at all? Thanks. ... For nonclustered indexes, absolutely! As only the index key is pushed into the upper levels of the index, the smaller the total key length, the better. ... Key Size Index Depth Total Page Count Fanout Average Time for selects Rough time per select.",
            "In my Insider newsletter a couple of weeks ago, I discussed how index fragmentation is often considered when designing indexes, but index depth often isn\u2019t. In the newsletter I said I\u2019d do a more comprehensive blog post with some data, so this is it. Fanout and Index Depth The index depth is determined by the fanout of [\u2026]"
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          "title": "How to figure out size of Indexes in MySQL - Stack Overflow",
          "url": "https://stackoverflow.com/questions/781873/how-to-figure-out-size-of-indexes-in-mysql",
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          "description": "If you are using InnoDB tables, you can get the size for individual indexes from mysql.innodb_index_stats. The &#x27;size&#x27; stat contains the answer, in pages, so you have to multiply it by the page-size, which is <strong>16K by default</strong>.",
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            "question": "How to figure out size of Indexes in MySQL",
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              "text": "<p>Extending Vajk Hermecz's answer.<br>\nThis is how you can get all of the indexes size, in megabytes, without the PRIMARY (which is the table itself), ordered by size.</p>\n\n<pre><code>SELECT database_name, table_name, index_name,\nROUND(stat_value * @@innodb_page_size / 1024 / 1024, 2) size_in_mb\nFROM mysql.innodb_index_stats\nWHERE stat_name = 'size' AND index_name != 'PRIMARY'\nORDER BY size_in_mb DESC;\n</code></pre>",
              "author": "Daniel Zohar",
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            "This gives the total index size, not a size per index (assuming a table has multiple indices). 2015-04-08T14:54:34.803Z+00:00 ... Save this answer. ... Show activity on this post. If you are using InnoDB tables, you can get the size for individual indexes from mysql.innodb_index_stats. The 'size' stat contains the answer, in pages, so you have to multiply it by the page-size, which is 16K by default.",
            "On MyISAM, each index block is 4 KB page filled up to fill_factor with index records, each being key length + 4 bytes long. ... As for InnoDB, the table is always clustered on the PRIMARY KEY, there is no separate PRIMARY KEY index ... Save this answer. ... Show activity on this post. Using phpMyAdmin, when viewing the table structure there is a Details link at the bottom somewhere. Once you click on it it will show you the total size of the indexes you have on the table where it is marked Space Usage.",
            "in my case it gave a very different (and unrealistic) size from the one reported by show table status from [dbname] 2017-07-28T07:41:56.413Z+00:00 ... Save this answer. ... Show activity on this post. Here's an adaption from some of the above to also give you the percentage of the total index for the table that each index has used, hopefully this will be useful for someone \u00b7 select database_name, table_name, index_name, round((stat_value*@@innodb_page_size)/1024/1024, 2) SizeMB, round(((100/(SELECT INDEX_LENGTH FROM INFORMATION_SCHEMA.TABLES t WHERE t.TABLE_NAME = iis.table_name and t.TABLE_SCHEMA = iis.database_name))*(stat_value*@@innodb_page_size)), 2) `Percentage` from mysql.innodb_index_stats iis where stat_name='size' and table_name = 'TargetTable' and database_name = 'targetDB'",
            "This is how you can get all of the indexes size, in megabytes, without the PRIMARY (which is the table itself), ordered by size. SELECT database_name, table_name, index_name, ROUND(stat_value * @@innodb_page_size / 1024 / 1024, 2) size_in_mb FROM mysql.innodb_index_stats WHERE stat_name = 'size' AND index_name != 'PRIMARY' ORDER BY size_in_mb DESC;"
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            "Similar to a book, the Database Engine writes all data rows on pages. The size of every page is the same: 8 KiB. In a book, most pages contain the data, or the main content of the book. Some pages contain metadata describing the content, for example, the table of contents and the index.",
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            "An index or partition has row-overflow data if the function returns rows where the alloc_unit_type_desc column is ROW_OVERFLOW_DATA and the page_count column is greater than 0. An extent is a collection of eight physically contiguous pages. The size of each extent is 64 KiB."
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