Mysql_num_rows not returning any value

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>";
   }
 }
?>

Consider using prepared statements to avoid SQL injection attacks. — Amber
– Amber, Commented Jan 26, 2015 at 6:52
What does the data look like in your database? are you sure the 'ximgid' you're using matches with one of the rows? — Tarske
– Tarske, Commented Jan 26, 2015 at 6:53
For delete query, use mysql_affected_rows() instead of mysql_num_rows() — Khushboo
– Khushboo, Commented Jan 26, 2015 at 6:53
This extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used. — Rafael
– Rafael, Commented Jan 26, 2015 at 6:56

Jens · Accepted Answer · 2015-01-26 06:54:45Z

4

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().

answered Jan 26, 2015 at 6:54

Jens

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3 Comments

Jens Over a year ago

@Khandaker Are you sure that records are deleted?

Jens Over a year ago

@Khandaker if you echoing $delete what is the value of that?

Khandaker Over a year ago

Nothing after echoing $delete , only blank page . @ Jens

Ian Thompson · Accepted Answer · 2015-01-26 07:38:59Z

0

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>";
}

answered Jan 26, 2015 at 7:38

Ian Thompson

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Comments

Khandaker · Accepted Answer · 2015-01-28 05:48:09Z

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.

Damian Nikodem · Accepted Answer · 2015-01-26 06:56:18Z

-1

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.

answered Jan 26, 2015 at 6:56

Damian Nikodem

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2 Comments

Muhammad Bilal Over a year ago

Please read question again, he said: delete query work properly

Damian Nikodem Over a year ago

oops, that is my bad, I missed it primarily because the English used in the post was not exactly clear so I jumped straight into looking at a fault in the code.

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