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From "Trejkaz (JIRA)" <j...@apache.org>
Subject [jira] Commented: (DERBY-3937) Select count(*) scans all the rows (and is therefore slow with big tables), is the amount of rows not available/known for example in index ?
Date Thu, 25 Jun 2009 04:35:07 GMT

    [ https://issues.apache.org/jira/browse/DERBY-3937?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=12723893#action_12723893
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Trejkaz commented on DERBY-3937:
--------------------------------

The query optimiser works on an estimation of the number of rows, not the actual number of
rows.

Related to this: in our case if we had a way to do APPROXCOUNT(*) instead of COUNT(*), it
would be fast and accurate "enough", as we are only using this on big tables to provide the
user with an idea of how much stuff is left to do.  It doesn't have to be the exact number
of items, just in the ballpark.  But we couldn't find a way to use the estimate, so we used
COUNT(*) and are thus bitten by this bug.


> Select count(*) scans all the rows (and is therefore slow with big tables), is the amount
of rows not available/known for example in index ?
> --------------------------------------------------------------------------------------------------------------------------------------------
>
>                 Key: DERBY-3937
>                 URL: https://issues.apache.org/jira/browse/DERBY-3937
>             Project: Derby
>          Issue Type: Improvement
>          Components: Performance
>         Environment: Any
>            Reporter: Martin Hajduch
>
> Create table with 5000000 rows. Create index on unique ID. Select count(*) on such table
is going to take quite some time.
> Shouldn't the index contain amount of indexed rows and the value taken from there ?
> Additionally, queries of the form select count(*) from table where col1=value; take lots
of time (depending on amount of rows satisfying WHERE clause) even if index on col1 exists.
Isn't it possible to find first and last occurence in the index, and then calculate amount
of rows more effectively then scanning through all of them ?

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