From: Gene Selkov, Jr. <selkovjr@mcs.anl.gov>
To: JT Kirkpatrick <jt-kirkpatrick@mpsllc.com>
To: pgsql-interfaces@hub.org
Subject: Re: [SQL] perl and postgres. . .
Date: Wed, 21 Apr 1999 15:36:08 -0400
Message-ID: <199904212036.PAA20785@antares.mcs.anl.gov> (raw)
In-Reply-To: <01BE8BF0.2DAA4CE0.jt-kirkpatrick@mpsllc.com>
> 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
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To: pgsql-interfaces@postgresql.org
Cc: selkovjr@mcs.anl.gov, jt-kirkpatrick@mpsllc.com, pgsql-interfaces@hub.org
Subject: Re: [SQL] perl and postgres. . .
In-Reply-To: <199904212036.PAA20785@antares.mcs.anl.gov>
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