Re: select IN problem - Mailing list pgsql-novice

From Andrew McMillan
Subject Re: select IN problem
Date
Msg-id 1014414921.3232.2399.camel@kant.mcmillan.net.nz
Whole thread Raw
In response to select IN problem  (Doug Silver <dsilver@quantified.com>)
Responses Re: select IN problem  (Doug Silver <dsilver@quantified.com>)
List pgsql-novice
On Sat, 2002-02-23 at 10:36, Doug Silver wrote:
> I've read the IN chapter in Bruce M.'s Postgresql book, but I still can't
> seem to get my select/IN to work.  I have two tables, transactions and
> transactions_detail, with the transaction_id field as the reference in
> the transactions_detail table.
>
> # select transaction_id from transactions where enter_date> cast('2002-02-22' as date);
>  transaction_id
> ----------------
>            2043
>            2044
>            2045
>
> purchases=# select transaction_id from transactions_detail where transaction_id>2042;
>  transaction_id
> ----------------
>            2043
>            2044
>            2045
>
> purchases=# \d transactions_detail
>             Table "transactions_detail"
>    Attribute    |         Type          | Modifier
> ----------------+-----------------------+----------
>  transaction_id | smallint              |
>  products       | character varying(20) |
>  quantities     | smallint              |
>
> But the following query causes it to hang, after 10 seconds I finally stop
> it.
>
> purchases=# select transaction_id from transactions_detail where
> purchases=# transaction_id IN (
> purchases=# select transaction_id from transactions where enter_date> cast('2002-02-22' as date)
> purchases=# );
>
> Any suggestions?

SELECT td.transaction_id FROM transactions_detail td
WHERE EXISTS (SELECT transaction_id FROM transactions t
   WHERE t.transaction_id = td.transaction_id
     AND t.enter_date > CAST('2002-02-02' AS DATE );

Could well work better.  The problem you are likely to be encountering
is that IN (...) will not use an index.

To see the query plans generated by the different SQL, use 'EXPLAIN <sql
command>' - it is _well_ worth coming to grips with what EXPLAIN can
tell you.

You could also be better with a plan that did a simple JOIN and
DISTINCT:

SELECT DISTINCT td.transaction_id
FROM transactions_detail td, transactions t
WHERE t.enter_date > '2002-02-02'
  AND td.transaction_id = t.transaction_id;

Regards,
                    Andrew.
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