Frequently Asked Questions (FAQ) for PostgreSQL
- Last updated: Mon Sep 30 23:28:35 EDT 2002
+ Last updated: Wed Oct 9 23:14:53 EDT 2002
Current maintainer: Bruce Momjian (pgman@candle.pha.pa.us)
4.22) Why are my subqueries using IN so slow?
Currently, we join subqueries to outer queries by sequentially
- scanning the result of the subquery for each row of the outer query. A
- workaround is to replace IN with EXISTS:
+ scanning the result of the subquery for each row of the outer query.
+ If the subquery returns only a few rows and the outer query returns
+ many rows, IN is fastest. To speed up other queries, replace IN with
+ EXISTS:
SELECT *
FROM tab
- WHERE col1 IN (SELECT col2 FROM TAB2)
+ WHERE col IN (SELECT subcol FROM subtab)
to:
SELECT *
FROM tab
- WHERE EXISTS (SELECT col2 FROM TAB2 WHERE col1 = col2)
+ WHERE EXISTS (SELECT subcol FROM subtab WHERE subcol = col)
- We hope to fix this limitation in a future release.
+ For this to be fast, subcol should be an indexed column. We hope to
+ fix this limitation in a future release.
4.23) How do I perform an outer join?
alink="#0000ff">
<H1>Frequently Asked Questions (FAQ) for PostgreSQL</H1>
- <P>Last updated: Mon Sep 30 23:28:35 EDT 2002</P>
+ <P>Last updated: Wed Oct 9 23:14:53 EDT 2002</P>
<P>Current maintainer: Bruce Momjian (<A href=
"mailto:pgman@candle.pha.pa.us">pgman@candle.pha.pa.us</A>)<BR>
<P>Currently, we join subqueries to outer queries by sequentially
scanning the result of the subquery for each row of the outer
- query. A workaround is to replace <CODE>IN</CODE> with
+ query. If the subquery returns only a few rows and the outer query
+ returns many rows, <CODE><SMALL>IN</SMALL></CODE> is fastest. To
+ speed up other queries, replace <CODE>IN</CODE> with
<CODE>EXISTS</CODE>:</P>
<PRE>
<CODE>SELECT *
FROM tab
- WHERE col1 IN (SELECT col2 FROM TAB2)
+ WHERE col IN (SELECT subcol FROM subtab)
</CODE>
</PRE>
to:
<PRE>
<CODE>SELECT *
FROM tab
- WHERE EXISTS (SELECT col2 FROM TAB2 WHERE col1 = col2)
+ WHERE EXISTS (SELECT subcol FROM subtab WHERE subcol = col)
</CODE>
</PRE>
+ For this to be fast, <CODE>subcol</CODE> should be an indexed column.
We hope to fix this limitation in a future release.
<H4><A name="4.23">4.23</A>) How do I perform an outer join?</H4>