Mysql_num_rows not returning any value - php

Question

What is the problem in my code ? every time it echo This image used as cover image , but delete query work properly.how can i fix it?
<?php
session_start();
if(!empty($_SESSION['userId']) && !empty($_SESSION['name'])){
include ('connect.php');
dbConnect();
if (isset($_GET['ximgid'])) {
$del=mysql_query("Delete FROM project_image WHERE ximgid='".$_GET['ximgid']."' And coverflag !='1'");
$delete=mysql_num_rows($del);
if($delete==1){
echo "<script type='text/javascript'>alert('Sucessfully Delete !!!')</script>";
echo "<script>javascript:window.location = 'projectImage.php'</script>";
}else{
echo "<script type='text/javascript'>alert('This Image Used As Cover Image')</script>";
echo "<script >window.location.href = 'projectImage.php'</script>";
}
}else{
echo "<script>javascript:window.location = 'index.php'</script>";
}
}
?>

Answer 1

mysql_num_rowsonly works with selectstatement.
For delete statements you have to use mysql_affected_rows
Here the part of the documentation of mysql_affected_rows:
Retrieves the number of rows from a result set. This command is only
valid for statements like SELECT or SHOW that return an actual result
set. To retrieve the number of rows affected by a INSERT, UPDATE,
REPLACE or DELETE query, use mysql_affected_rows().

Answer 2

mysql_affected_rows() is the way to go
$delete=mysql_affected_rows($del);
if($delete>0){
echo "<script type='text/javascript'>alert('Sucessfully Delete !!!') </script>";
echo "<script>javascript:window.location = 'projectImage.php'</script>";
}else if($delete==0){
echo "<script type='text/javascript'>alert('This Image Used As Cover Image')</script>";
echo "<script >window.location.href = 'projectImage.php'</script>";
}else{
echo "<script type='text/javascript'>alert('mysql error')</script>";
}

Answer 3

I had to solve my problem in an alternative way which is given below:
<?php
session_start();
if(!empty($_SESSION['userId']) && !empty($_SESSION['name'])){
include ('connect.php');
dbConnect();
if(isset($_GET['ximgid'])){
$selQuery = "SELECT coverflag FROM project_image WHERE ximgid ='".trim($_GET['ximgid'])."' AND zid = '1000' AND xpid = '".trim($_GET['pid'])."'";
$flagtResult=mysql_query($selQuery);
$flagRow=mysql_fetch_array($flagtResult);
mysql_free_result($flagRow);
$cover=$flagRow[0];
if($cover == 1){
echo "<script type='text/javascript'>alert('Used as cover picture')</script>";
echo "<script>javascript:window.location = 'projectImage.php'</script>";
}else{
$sql = "DELETE FROM project_image WHERE ximgid = '".trim($_GET['ximgid'])."' AND coverflag != '1' AND zid = '1000' AND xpid = '".trim($_GET['pid'])."'";
$result = mysql_query($sql) or die(mysql_error());
if($result > 0){
echo "<script type='text/javascript'>alert('Sucessfully Detele !!!')</script>";
echo "<script>javascript:window.location = 'projectImage.php'</script>";
}
}
}
}else{
echo "<script>javascript:window.location = 'index.php'</script>";
}
?>
Now I can get my proper output.. Thanks Everyone.

Answer 4

first of all it appears that you are escaping ximgid and coverflag with ' which would throw a syntax error in your sql if those were not varchar fields.
$del=mysql_query("Delete FROM project_image WHERE ximgid=".$_GET['ximgid']." And coverflag !=1;");
Second issue to address would be that you are taking a $_GET parameter and putting it directly into your SQL query which is just trouble looking to happen so I would suggest next changing it to:
$del=mysql_query("Delete FROM project_image WHERE ximgid=".mysql_real_escape_string($_GET['ximgid'])." And coverflag !=1;");
if you need extra help debugging your SQL you can always temporarily put a call to
echo mysql_error();
after your call to mysql_query in order to show the error message from the server.