I am finding what is the best way to estimate size of a table for that i have studied lot of blogs and forums but unable to find any accurate answer

For an example we have a table City with InnoDB engine,lets say in future (in next 1 year) it will have 1 million of records so what will be the estimated data size and index size of that table in this period.

mysql> desc City;
| Field       | Type     | Null | Key | Default | Extra          |
| ID          | int(11)  | NO   | PRI | NULL    | auto_increment |
| Name        | char(35) | NO   |     |         |                |
| CountryCode | char(3)  | NO   | MUL |         |                |
| District    | char(20) | NO   |     |         |                |
| Population  | int(11)  | NO   |     | 0       |                |
5 rows in set (0.03 sec)


What will be the estimated upper bound (Max size of table) with 1 million records and how can we estimate it.

  • This is great. but is it possible to get the index size column wise. What is mean is if you have a table with (say) 5 columns indexed. Can we get the index size of each one of them? I will ask this as another question. thanks – Sushil Sep 11 '13 at 10:04

Given the table description, I see

  • 66 bytes per row of data
  • 4 bytes per row for the primary key
  • 7 bytes per row for country code index
    • 3 bytes for the country
    • 4 bytes for Clustered Key attached to the country code
  • Total of 77 bytes of data and keys
  • This does not factoring housekeeping for BTREEs or Tablespace Fragmentation

For a million rows, that would 77,000,000 bytes (73.43 MB)

As for measuring the table, for a given table mydb.mytable, you can run this query

    CONCAT(FORMAT(DAT/POWER(1024,pw1),2),' ',SUBSTR(units,pw1*2+1,2)) DATSIZE,
    CONCAT(FORMAT(NDX/POWER(1024,pw2),2),' ',SUBSTR(units,pw2*2+1,2)) NDXSIZE,
    CONCAT(FORMAT(TBL/POWER(1024,pw3),2),' ',SUBSTR(units,pw3*2+1,2)) TBLSIZE
    SELECT DAT,NDX,TBL,IF(px>4,4,px) pw1,IF(py>4,4,py) pw2,IF(pz>4,4,pz) pw3
        SELECT data_length DAT,index_length NDX,data_length+index_length TBL,
        FLOOR(LOG(IF(data_length=0,1,data_length))/LOG(1024)) px,
        FLOOR(LOG(IF(index_length=0,1,index_length))/LOG(1024)) py,
        FLOOR(LOG(data_length+index_length)/LOG(1024)) pz
        FROM information_schema.tables
        WHERE table_schema='mydb'
        AND table_name='mytable'
    ) AA
) A,(SELECT 'B KBMBGBTB' units) B;

To measure all tables grouped by Database and Storage Engine

    IF(ISNULL(DB)+ISNULL(ENGINE)=2,'Database Total',
    CONCAT(DB,' ',IFNULL(ENGINE,'Total'))) "Reported Statistic",
    LPAD(CONCAT(FORMAT(DAT/POWER(1024,pw1),2),' ',
    SUBSTR(units,pw1*2+1,2)),17,' ') "Data Size",
    LPAD(CONCAT(FORMAT(NDX/POWER(1024,pw2),2),' ',
    SUBSTR(units,pw2*2+1,2)),17,' ') "Index Size",
    LPAD(CONCAT(FORMAT(TBL/POWER(1024,pw3),2),' ',
    SUBSTR(units,pw3*2+1,2)),17,' ') "Total Size"
    IF(px>4,4,px) pw1,IF(py>4,4,py) pw2,IF(pz>4,4,pz) pw3
    (SELECT *,
        FLOOR(LOG(IF(DAT=0,1,DAT))/LOG(1024)) px,
        FLOOR(LOG(IF(NDX=0,1,NDX))/LOG(1024)) py,
        FLOOR(LOG(IF(TBL=0,1,TBL))/LOG(1024)) pz
        SUM(data_length) DAT,
        SUM(index_length) NDX,
        SUM(data_length+index_length) TBL
       SELECT table_schema DB,ENGINE,data_length,index_length FROM
       information_schema.tables WHERE table_schema NOT IN

Run these queries and you can track changes in database/engine disk usage.

Give it a Try !!!

| improve this answer | |
  • 1
    This is a really great query for viewing all of your table sizes. – ghayes Apr 15 '14 at 22:01
  • The CHAR lengths need to be multiplied by 3 if you have CHARSET utf8. The entire overhead can be estimated by doubling or tripling the computation. – Rick James Aug 18 '15 at 23:27
  • @RolandoMySQLDBA , do you know if is possible to calculate the "real" row size of a table with the objective to compare with the real size (compressed table) and get the ratio of compress? – ceinmart Jun 19 '19 at 20:26
  • @ceinmart innodb_page_size is fixed (16K or 16384 by default) and becomes the boundary where rows and grouped or split. Changing innodb_page_size can alter storage of data for good or bad.Based on the how filled or sparse a row is (especially with the presence of TEXT/BLOB/VARCHAR). At best, you should compare the the size of .ibd file to what the schema report to estimate a ratio. You may also need to perform a NULL ALTER TABLE (ALTER TABLE ... ENGINE=InnoDB;)to get an accurate ratio. Effort may not be worth it. – RolandoMySQLDBA Jun 19 '19 at 21:13
  • @ceinmart Keep in mind hat changing innodb_page_size is not a table-by-table setting. You would need to do a full export of the data (See mariadb.com/kb/en/library/how-to-change-innodb_page_size) – RolandoMySQLDBA Jun 19 '19 at 21:15

If you are using InnoDB tables, you can get the size for data/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, that is 16K by default.

select database_name, table_name, index_name, stat_value*@@innodb_page_size
from mysql.innodb_index_stats where stat_name='size';

The index PRIMARY is the data itself.

| improve this answer | |
  • 1
    This assumes you have data in the table; seems like the OP wants to estimate before populating. – Rick James Nov 12 '19 at 6:25

If you don't have data yet, here are some tips. The following applies to InnoDB. (MyISAM is much simpler, and smaller.)

Don't use CHAR for variable-length columns. What CHARACTER SET are you using? Ascii needs one byte per character; utf8mb4 needs between 1 and 4.

4 bytes per INT
35 for CHAR(35), if ascii or latin1; varchar is probably less
3 for the country code; it is fixed length

Total = about 80 bytes.

Multiply the 80 by between 2 and 3 to account for various overheads. Most likely the 1M row table will be between 160MB and 240MB.

To measure a single index, for say CountryCode of 3 bytes:

3 bytes data
4 bytes for the PK (implicitly included with any secondary key)
25 bytes basic overhead
32 total
times 1.5 -- overhead for BTree that was randomly inserted into
48MB -- total for 1M rows.


  • Only the leaf nodes (of BTrees) need to be computed; the overhead for the non-leaf nodes is typically 1%.

  • The PRIMARY KEY is "clustered" with the data, so there is no need to compute it.

  • If you do not have an explicit PK, then you need to add 6 bytes to the row size to allow for the fabricated PK.

  • ROW_FORMAT = COMPRESSED gives you about a 2:1 shrinkage. (This is not as good as typical zip (etc) compression rate of 3:1.)

  • SHOW TABLE STATUS LIKE "tablename"; is the quick way to compute the 'actual' size. See Data_length for data and PK; Index_length for secondary indexes, and Data_free for some other stuff.

  • It is rare for Index_length to exceed Data_length. However it is not "wrong" for that to happen.

| improve this answer | |
SELECT  Table_NAME "tablename",
           data_length   "table data_length in Bytes",
           index_length  "table index_length in Bytes",
           data_free  "Free Space in Bytes"
    FROM  information_schema.TABLES  where  Table_schema = 'databasename';

by executing this query you can get size used for Data and Index of a table , You can check this size against # of rows and predict for 1 million rows

| improve this answer | |
  • 1
    I am not sure but will this give some what accurate results ? have you tested this ever ? – Abdul Manaf Jul 10 '13 at 12:18
  • Actually I am testing this query result periodically to see growth (%) w.r.t. size – Peter Venderberghe Jul 10 '13 at 12:40

It's tedious. But the details are in the docs.

To be as accurate as possible, which is rarely necessary, you'll need to read about the table structure and index structure, too.

If I were in your shoes, I'd build the table, populate it with a million rows of test data, and measure the change in size. Depending on your application, you might need to take the size of transaction log files into account, too.

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