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up vote 0 down vote favorite

I can't seem to get past this error.

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /windsor.php on line 47

if ($insertdb == NULL)
  mysql_query("INSERT INTO `asd` (`id`, `1`, `2`,`3`, `4`, `5`, `pubdate`) VALUES (296, maddeal', 'Windsor', 'ON', '', '', '') ON DUPLICATE KEY UPDATE `1`=VALUES(`1`),`2`=VALUES(`2`),`3`=VALUES(`3`),`4`=VALUES(`4`),`5`=VALUES(`5`)") or die(mysql_error());
 else
//Check and see if value has changed...

$checksql = mysql_query("SELECT `1` FROM deal WHERE `id`= 296") or die(mysql_error());
while($row = mysql_fetch_array($checksql))
  {
  $checksqlver = $row['deal'];
  }

$checksqlver = mysql_real_escape_string($checksqlver);
//$checksqlver = stripslashes($checksqlver);
echo "$checksqlver<br>";

if ($checksqlver == $insertdb)
  exit();
 else
  echo "No Match<br>";

//percent
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2 Answers 2

up vote 1 down vote accepted

Did you connect to the database using mysql_connect?

That would cause the error you are describing.

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Yes, I have that specified at the top, just didn't post. Other scripts work, just this one is giving me grief... –  mrlayance Jan 28 '11 at 1:05
    
You also have an else statement there where you don't have any code until the $checksql = ... try removing that else. –  Alec Gorge Jan 28 '11 at 1:08
    
That was it... kinda –  mrlayance Jan 28 '11 at 1:21
    
replaced else //Check and see if value has changed... with else echo "Moving On"; Not that best fix, but it works. –  mrlayance Jan 28 '11 at 1:22
    
Excellent. PHP could really stand to make their errors a little more specific as to what the problem actually is... In your case it is because you passed a null value to mysql_fetch_array because $checksql was declared in a different scope. –  Alec Gorge Jan 28 '11 at 1:23
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up vote 0 down vote

In your select query should 1 have apostrophes around it?

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