Re: Var substitution in SELECT statements

On Mon, Apr 23, 2001 at 09:41:51PM -0400, Randall Perry wrote:
> on 4/23/01 9:20 PM, Randall Perry at rgp(at)systame(dot)com wrote:
>
> > This works:
> > $res = $conn->exec("select cust, contact, user_name, email from $t where
> > user_name = a1a");
> >
> > This doesn't:
> > $c = "a1a";
> > $res = $conn->exec("select cust, contact, user_name, email from $t where
> > user_name = $c");
> >
> > and returns the error:
> > Attribute 'a1a' not found
> >
> >
> > How do you do var substitution with the Pg module in Perl?
>
> Whoops! Needed to quote the var as so (for $c above):
>
> $c = "\'$c\'";
>
> Works now.

also try

$sth = $dbh->prepare("select fields from tbl where f1 = ? and f2 = ?")
$sth->execute($val1,$val2);
while ($ref = $sth->fetchrow_hashref()) {
...
}
$sth->finish();

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