PHP mysql_fetch_array Issue

Question

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

Alec Gorge · Accepted Answer · 2011-01-28 00:59:51Z

1

Did you connect to the database using mysql_connect?

That would cause the error you are describing.

answered Jan 28, 2011 at 0:59

Alec Gorge

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4 Comments

mrlayance Over a year ago

Yes, I have that specified at the top, just didn't post. Other scripts work, just this one is giving me grief...

Alec Gorge Over a year ago

You also have an else statement there where you don't have any code until the $checksql = ... try removing that else.

mrlayance Over a year ago

replaced else //Check and see if value has changed... with else echo "Moving On"; Not that best fix, but it works.

Alec Gorge Over a year ago

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.

Trinitrotoluene · Accepted Answer · 2011-01-28 01:01:48Z

0

In your select query should 1 have apostrophes around it?

answered Jan 28, 2011 at 1:01

Trinitrotoluene

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