mysql_num_rows validation problem

[s2e]

This is a very simple query, but I am not seeing my problem.
Using GET, I am pulling the url to a pic from the database. If no url is present in the db (ie no pictures have been uploaded) then the script should echo 'Photo not on file'. Seems that I have a problem with the validation I am trying to use. I want to simply count the number of rows obtained by the query (which should be either 0 or 1) and based on this, echo either the picture or the message.

Here is my code:

<?php
$mem_id = $_GET['mem_id'];
require_once('./mysql_connect.php');
$query = "SELECT picture1 FROM memorial WHERE mem_id = '$mem_id'";
$result = @($query);
$check = mysql_num_rows($result);
if ($check > 0) {
$row = @mysql_fetch_array($result, MYSQL_NUM);
$pic = $row[0];
echo "<img src=$pic>";
} else {
echo "No Photo on File";
}
?>

No matter what, the first condition is always echoed and if there is no picture uploaded, I get a blank image instead of the 'Photo not on file' message.
Any Help?

[coreyp_1]

how about

<?php
  $mem_id = $_GET['mem_id'];
  require_once('./mysql_connect.php');
  $query = "SELECT picture1 FROM memorial WHERE mem_id = '$mem_id'";
  $result = @mysql_query($query);
  if (mysql_num_rows($result)) {
    $row = @mysql_fetch_array($result, MYSQL_NUM);
    $pic = $row[0];
    echo "<img src=$pic>";
  } else {
    echo "No Photo on File";
  }
?>

this is more direct, and doesn't waste a variable.

if this doesn't work, consider the possibility that there actually is a record for this mem_id in the table, but it's an empty string. To exclude this possibility, you will need to add an additional WHERE clause.

since you are only looking for one row, you might also consider adding LIMIT 1 to your query.

[s2e]

Thanks coreyp_1. Getting the same results though.
Added 'X' by default in place of empty fields in the db so i am not counting, or working with any empty fields, and tried this...

<?php
$mem_id = $_GET['mem_id'];
require_once('./mysql_connect.php');
$query = "SELECT picture1 FROM memorial WHERE mem_id = '$mem_id' LIMIT 1";
$result = @($query);
$row = @mysql_fetch_array($result, MYSQL_NUM);
$pic = $row[0];
if($pic = 'X') {
echo 'No Photo On File';
} else {
echo "<img src=$pic><hr size=1>";
}
?>

This returns the 'No Photo On File' message every time. Seems like in each case it returns the first echo command regardless of the if/else statement.
Any Ideas?

[batman]

Why not just do this:

<?php
$mem_id = $_GET['mem_id'];
require_once('./mysql_connect.php');
$query = "SELECT picture1 FROM memorial WHERE mem_id = '$mem_id' LIMIT 1";
$result = @mysql_query($query);
if(!$result)
{
echo 'No Photo On File';
} 
else 
{
while($row = mysql_fetch_array($result))
{
$pic = $row['picture1'];
}
echo "<img src=\"$pic\"><hr size=1>";
}

[s2e]

Yes that code looks good too but still producing the same results.
Echoing a blank image and the horizontal rule even when no url is present. Don't understand why its ignoring the if statement. Getting very late though. Thanks everyone for the help. I'll try again tomorrow.