RE: php & mysql prob...

From: Date: Wed, 31 Oct 2001 09:30:07 +0000
Subject: RE: php & mysql prob...
Groups: php.general 
Request: Send a blank email to php-general+get-72995@lists.php.net to get a copy of this message
Your query is invalid, hence $test is false rather than a result set. I usually do if ($test = mysql_db_query( ... )) to trap this. Don't put quotes around table or field names, in your case 'Port$p' should be just Port$p and 'Port$p'.date just Port$p.date. Tim ---------- From: sc [SMTP:sven@gstar.com.au] Sent: 31 October 2001 00:56 To: php-general@lists.php.net Subject: php & mysql prob... Hi; i keep getting an error of: Warning: Supplied argument is not a valid MySQL result resource in /datascripts/insertdata.php on line 17... Line 17 is: $row = mysql_fetch_assoc($test); and here is the rest of it (not all of it though): for ($p = 1; $p <= 24; $p++) { $test = mysql_db_query("melbourne", "SELECT * FROM 'Port$p' WHERE 'Port$p'.date='$yesterday'"); $row = mysql_fetch_assoc ($test); $yindata = $row['switchin']; $youtdata = $row['switchout']; $dinPort = '$inPort$p' - $yindata; $doutPort = '$outPort$p' - $youtdata; Can anyone help me overcome this prob? i've prob missed something without thinking but i cant seem to get it... Thx. sc

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