Im using PHP & MySQL, the Problem im facing to INSERT data using "mysql_fetch_array".
this is my my connection to Mysql and my query to display data.
There are 2 table in this. 1 table for display. 1 more for insert data.
<?php
$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password=""; // Mysql password 
$db_name="skpj"; // Database name  
// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM `student` WHERE cls_id = '13' ";
$result=mysql_query($sql);
// Count table rows 
$count=mysql_num_rows($result);
?>    <form name="form1" method="post" action="">
    <?php while($rows=mysql_fetch_array($result)){ ?>
    <?php echo $rows['s_no']; ?>
    <?php echo $rows['name']; ?>
    <input name="s_no[]" type="hidden" id="name" 
value="<?php echo $rows['s_no']; ?>">
    <?php echo $rows['ic']; ?>
    <?php echo $rows['cls_id']; ?>
    <input name="class_n[]" type="hidden" id="cls_id" 
value="<?php echo $rows['cls_id']; ?>">
    <select name="att[]" id="att" style=" width:80px" >
      <option value="1">Atten</option>
      <option value="2">Absend</option>
      <option value="3">MC</option>
    </select>
    <input name="tmp[]" type="hidden" id="name" value="1">
    <?php } ?>
    <input type="submit" name="submit" value="submit"></td>
</form>
<?php
if($submit){
   for($i=0;$i<$count;$i++){
      $sql1="INSERT INTO attendance (s_no, class_n, att, tmp) 
      VALUE '$s_no[i]','$class_n[i]','$att[i]','$tmp[i]' ";
      $result1=mysql_query($sql1);
   }
}
if($result1){
  header("location:att2.php");
}
mysql_close();
?>below is the notice foe my error.
SCREAM: Error suppression ignored for
Notice: Undefined variable: submit in C:\wamp\www\att\att2.php on line 67
Notice: Undefined variable: result1 in C:\wamp\www\att\att2.php on line 74
furthermore i cannot insert my data
 
     
     
    