Warning: mysql_num_rows(): supplied argument Error

[Skor]

I'm receiving the error above in my shopping cart application.  I sell one-off items and use Paypal as the payment engine.  In the remote possibility an item was previously purchased I loop through each order and update one at a time.  I'm counting rows so if a record is not updated, I'd like to know.  Strangely, it's working as designed (sending emai), but don't want to deploy unless I know why this error is occuring.

 

Here's what I've got so far.  Here's the full error:

 

1.    is not a valid MySQL result resource in cartworks15.php on line 105.

 

2. Do I need to  change the variable to a different function.  Is my syntax correct.

 

Please let me know.  Thanks in advance.

 

3. The code:

 

mysql_query ($query);

// breaks the string into an array of products
$products = explode("~", $cart);

$i = 0;

// breaks each product string into its own array
foreach($products as $var => $val){

  $product[$i] = explode(":", $val);
  $desc = $product[$i][0];
  $qty = $product[$i][1];
  $money = $product[$i][2];
  $pid = $product[$i][4];

$update = "UPDATE ckd_labinv SET status ='S',
  sold_code = 'COD', sold_date = NOW()
  WHERE pid = '$pid'
  and status = 'A'";
  mysql_query($update) or die(mysql_error());

  $i++;
}

//echo '...Updated Database <br>'; 

$to = "[email protected]";
$subject = "Already Purchased Alert!";
$contents = "\nThe following item has already been purchased:\n\n * $desc \n\nThis item was purchased by the following customer:\n\n  $inv_name,$email, on $date.\n";
$from_header = "From: [email protected]";
$num_rows = mysql_num_rows($update);  [color="Red"][b]//offending line[/b][/color]
//mysql_affected_rows($update);

if ($num_rows == 0) 
{
mail( $to, $subject, $contents, $from_header);
echo '...sent Already Purchased Alert <br>'; 
}

 

 

Thanks in advance.

[trq]

This is a common error when your query has failed and you have not checked your queries result. At the very least you should always use...

 

mysql_query ($query) or die(mysql_error());

 

But (IMO) its best to use....

 

<?php

  if ($result = mysql_query($sql)) {
    // $result is safe to use.
    if (mysql_num_rows($result)) {
      // $result holds records.
    } else {
      // no records found.
    }
  } else {
    // query failed, handle error gracefully.
  }

?>