Re: [GENERAL] Var substitution in SELECT statements

2001-04-24 Thread will trillich
On Mon, Apr 23, 2001 at 09:41:51PM -0400, Randall Perry wrote: > on 4/23/01 9:20 PM, Randall Perry at [EMAIL PROTECTED] wrote: > > > This works: > > $res = $conn->exec("select cust, contact, user_name, email from $t where > > user_name = a1a"); > > > > This doesn't: > > $c = "a1a"; > > $res =

Re: [GENERAL] Var substitution in SELECT statements

2001-04-23 Thread Randall Perry
on 4/23/01 9:20 PM, Randall Perry at [EMAIL PROTECTED] 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 = $

[GENERAL] Var substitution in SELECT statements

2001-04-23 Thread Randall Perry
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