Re: [SQL] Update problem I cannot work out - Mailing list pgsql-sql

From Tom Lane
Subject Re: [SQL] Update problem I cannot work out
Date
Msg-id 13436.925663017@sss.pgh.pa.us
Whole thread Raw
In response to Re: [SQL] Update problem I cannot work out  (M Simms <grim@argh.demon.co.uk>)
List pgsql-sql
M Simms <grim@argh.demon.co.uk> writes:
>> Create function new_function(value_data_type)
>> 'Select sum(t1.var1) from t1 where t1.var2 = $1'  LANGUAGE
>> 'sql';
>> 
>> update t2 set var1=t2.var1+new_function(t2.var2);

> My reply to this is the question, that as each function contains a
> select, is this technically a single pass.

I think you are right: this way will repeat the sub-select for each
tuple in t2.  Now if you have an index on t1.var2, the system could
exploit it to pull out just the desired tuples in each sub-select, so
the total work wouldn't really be O(M*N).  But it still looks slow.

If I knew t1 would be large, I think I'd do it with a temporary table:
SELECT var2, sum(var1) FROM t1 GROUP BY var2 INTO temptable;UPDATE t2 SET var1 = t2.var1 + temptable.sum    WHERE
t2.var2= temptable.var2;DROP TABLE temptable;
 

(In 6.5 you can use a temporary table to avoid needing explicit DROP,
but that's just a convenience.)  Perhaps there's a way to accomplish
the same thing with a subselect, rather than explicitly creating a
table, but I don't know how.

This is more like O(N log N) in the size of each table, assuming that
the join is done with mergejoin.  Actually, if there's not a huge
number of distinct values of var2, then temptable should be small enough
for a hashjoin, which'd mean that there's no need to sort t2 at all.

BTW, I think your original try

update t2 set var1=t2.var1+t1.var1 where t2.var2=t1.var1;

should have worked; it's probably a bug that it doesn't work.  IMHO
the semantics ought to be that the addition is repeated for each t1
tuple that matches the current t2 tuple ... but I haven't read the
SQL spec.
        regards, tom lane


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