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

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""lameck kassana"" <chelaskk@xxxxxxxxx> wrote in message 
news:967cd34f0707110248w50d29f31hde718e66e54c9e2a@xxxxxxxxxxxxxxxxx
> 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 

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