Re: Join causing Error?

From: Date: Sun, 25 Feb 2001 09:12:41 +0000
Subject: Re: Join causing Error?
References: 1  Groups: php.general 
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In article <030601c09f05$2b9e6e70$53daead8@gilanet.com>, larentium@gilanet.com ("Keith Spiller") wrote: > Can anyone tell me why this: > Line 282 mysql select db("centraldb",$db); > Line 283 $qorder++; > Line 284 $result = mysql query("SELECT q.questid, q.question, q.answer, > q.qorder, q.depart, q.catid, > q.active, q.global, q.adate, q.author, q.authoremail, > q.askemail, c.catid, c.category, c.under, > c.corder, c.active FROM central groupfaqq q, central > groupfaqcat c WHERE q.active = '1' AND > q.global = '1' AND c.active = '1' ORDER BY c.under, > c.order, q.qorder",$db); > Line 285 while ($myrow = mysql fetch row($result)) > > Would cause this error: > Warning: Supplied argument is not a valid MySQL result resource in > faqbody.php3 on line 285 > > When changing the same SELECT statement to: > Line 284 $result = mysql query("SELECT * FROM central groupfaqq WHERE > active = '1' ORDER BY > qorder",$db); > > Works perfectly? Sounds like the 'while' loop is failing because mysql_query isn't returning to $result what you think it is. Add some error-checking, starting with at least an "or die(mysql_error())" on the query. Also, try running the same query at the command line to confirm that it's valid. -- CC

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