Date: Tue, 24 Apr 2001 02:15:16 -0500
From: will trillich <will@serensoft.com>
To: pgsql-general@postgresql.org
Subject: Re: Var substitution in SELECT statements
Message-ID: <20010424021516.I27356@serensoft.com>
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References: <B70A4D23.A28D%rgp@systame.com> <B70A521F.A291%rgp@systame.com>
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In-Reply-To: <B70A521F.A291%rgp@systame.com>;
	from rgp@systame.com on Mon, Apr 23, 2001 at 09:41:51PM -0400

On Mon, Apr 23, 2001 at 09:41:51PM -0400, Randall Perry wrote:
> on 4/23/01 9:20 PM, Randall Perry at rgp@systame.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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will@serensoft.com
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