Empty db field, if statement never checks for empty

kilbey1

I've got a small bit of code here that never checks for an empty value (in other words, it always finds something in there, even if it's completely empty in the database). Can anyone else see my problem?

If there is an entry in the DB named default.gif or some other image, and they've chosen to upload nothing, keep the image as is. But if it's empty in the database (and they've still chosen to not upload an image), we need to put in default.gif.

	$sql_image = "SELECT picture, product_id 
	FROM products
	WHERE product_id='$pid'";
	$query_image = mysql_query ($sql_image, $db_connection) or die (mysql_error());
	$query_row = mysql_fetch_array ($query_image);

if (($query_row[picture] != 'none') || ($query_row[picture] != 'default.gif')) {
		echo "Image already exists. Keeping current image.<br>";
		$userfile_name = $query_row[PIC];
} elseif ($query_row[picture] == 'none' || '' || 'empty' || 'NULL') {
		echo "No image exists. Reverting to default image.<br>";
		$userfile_name = "default.gif";
}

diego25

You cannot do

} elseif ($query_row[picture] == 'none' || '' || 'empty' || 'NULL') {

You have to write $query_row['picture'] every single time, e.g.

} elseif ($query_row['picture'] == 'none' || $query_row['picture'] == '' || etc..

And please, use quotes around 'picture'. You should turn your error_reporting to E_ALL, so you see all the errors.

Diego

kilbey1

Hey, thanks for the tip about e_all, I never knew about that. Hey, I'm still learning. 🙂

After your changes, I had to switch around the checking - first for empty, then for an existing image. Here we go:

	$sql_image = "SELECT picture, product_id 
	FROM products
	WHERE product_id='$pid'";
	$query_image = mysql_query ($sql_image, $db_connection) or die (mysql_error());
	$query_row = mysql_fetch_array ($query_image);

if ($query_row['picture'] == 'none' || $query_row['picture'] == '' || $query_row['picture'] == 'empty' || $query_row['picture'] == 'NULL') {
		echo "No image exists. Reverting to default image.<br>";
		$userfile_name = "default.gif";
}
else if ($query_row['picture'] != 'none' || $query_row['picture'] != 'default.gif') {
		echo "Image already exists. Keeping current image.<br>";
		$userfile_name = $query_row['picture'];
}

Thanks again!
- Eve