Обсуждение: Var substitution in SELECT statements
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? -- Randy Perry sysTame Mac Consulting/Sales
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. -- Randy Perry sysTame Mac Consulting/Sales
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(); -- don't visit this page. it's bad for you. take my expert word for it. http://www.salon.com/people/col/pagl/2001/03/21/spring/index1.html will@serensoft.com http://sourceforge.net/projects/newbiedoc -- we need your brain! http://www.dontUthink.com/ -- your brain needs us!