Re: Newbie alert: supplied argument is not a valid MySQL result resource

""lameck kassana"" <chelaskk@gmail.com> wrote in message news:967cd34f0707110248w50d29f31hde718e66e54c9e2a@mail.gmail.com... > hey just make this > $sid = mysql_fetch_array($query) or die ('Could not fetch from database: >> ' ." mysql_error()."); > try this Yep. I figured that out yesterday. Hit the nail right on the head. As I've learned from experimentation, the pipe operator takes precedence to the assignment operator so I was basically evaluating whether my query or a die statement returned a true value (which the query statement was, so it never processed the die()) and then assigning "true," or 1 to $query. The 'or' operator by contrast processes after the assignment, so the following line: $query = mysql_query("some SQL") or die(); first assigns a handle to $query (assuming the query is valid, which in my case it was) and then evaluated the value of query. This leaves me with two remaining questions. I'm coming from a Perl background, so I'm wondering if there's a documentation page somewhere for PHP operators (something along the likes of perlop) and as an experiment I tried the following code, and it behaved exactly like the code I first posted, which I'm not sure why: ($query = mysql_query("valid SQL")) || die(); Shouldn't everything within the first set of parentheses here evaluate as the first || condition? Thanks for everyone's help, Matt