Re: [SQL] perl and postgres. . .
От | Gene Selkov, Jr. |
---|---|
Тема | Re: [SQL] perl and postgres. . . |
Дата | |
Msg-id | 199904212036.PAA20785@antares.mcs.anl.gov обсуждение исходный текст |
Список | pgsql-interfaces |
> I am trying > $query="select max((userseq) from dataentry;"; > $result=$conn->exec("$query"); > $userseq=($result->getvalue(0,0)); > > * I need the value in a variable name other than $result. . . > > If I try > $query="select max((userseq) from dataentry;"; > $result->getvalue(0,0); > $userseq=$result; > > it bombs. Ahhh, the frustrations of being new to this! Now I see what your problem is. You missed the whole point. $result is not a variable, it is an object reference, so is $conn. The exec() method of $conn (or rather, that of the object referenced by $conn) returns the result object. You normally access the object's data through its methods, in this case, getvalue(), fetchrow(), etc. -- see the Pg doc for the full list. So the following simply copies the reference to the result object into $userseq, and that may bomb, depending on how you use it further: > $userseq=$result; The following the value you need, but it since there is no explicit assignment, it is getting lost: > $result->getvalue(0,0); What you needed was $userseq = $result->getvalue(0,0); And again, the manual for Pg has complete examples. Note the code you need to add to check for errors that may result from each method call. --Gene
В списке pgsql-interfaces по дате отправления: