Question

I want to display values retrieved from mysql in forms and in row by row format.
So i used the following code.

<?php 
  while($rows1 = mysql_fetch_array($result1)){ 
?>
<tr height="30" > 
<?php $elyid = $rows1['id'] ?>
  <form action="leaveactions.php" method="post" name="viewleave">
   <td width="82"><?php echo $rows1['empid'];?></td>
   <td><?php echo $rows1['name'];?></td>
   <td><?php echo $rows1['leavetype'];?></td>
   <td width="82"><?php echo $rows1['startdate']; ?></td>
   <td width="82"><?php echo $rows1['enddate']; ?></td>
   <td><?php echo $rows1['leavetype']; ?></td>
   <td>
     <input type="submit" name="<?php $rows1['id']; ?>" value="accept"/>
   </td>
   <td>
     <input type="submit" name="reject" value="reject"/>
     <input type="hidden" name="emplid" value="<?php echo $rows1['id'] ?>"/>
   </td>
</tr>
<?php       } ?>

and in leaveactions.php

$ii=0;
$query1 = "select * from applied_leaves where supervisorid ='".$employeeId."' and status='not approved'";
$result1 = mysql_query($query1) or die (mysql_error());
$num1 = mysql_numrows($result1);

while($rows1 = mysql_fetch_array($result1)) {
  $ii++;
  echo $_POST["$ii"];
  if(isset($_POST['$ii'])){
    echo "accepted "; echo $_POST['$ii'];
    $updateEmp = "update applied_leaves set status='".$accept."' where id='$ii' " ;
    $uresult = mysql_query($updateEmp) or die (mysql_error());
    if($uresult != null){
      echo "Assignment Added successfully<br>";
?>
      <a href="updateassignment.php">View Added Details</a>
<?php                       
    } else {
      echo "error";
    }
  }
}
?>

but when run i get

 Notice: Undefined offset: 1
 Notice: Undefined offset: 2
.
.
.
.
.
.

like that.
Please help me solving the problem.
Thanks in Advance

EDIT

the new code that is causing exception is

                     $iii=0;
                while($rows1 = mysql_fetch_array($result1))
                {
                  $iii++;

                if(  $_POST["accpt".$iii] ) { 

                  echo "accepted "; 
                  $updateEmp = "update applied_leaves set status='".$accept."' where id='$iii' " ;
                  $uresult = mysql_query($updateEmp) or die (mysql_error());
                  if($uresult != null){
                        echo "Assignment Added successfully<br>";
                        ?>
                        <a href="updateassignment.php">View Added Details</a>
                        <?php   break;                  
                    }

                }

                }

Answer 1

I think your main error lies here:

<input type="submit" name="<?php $rows1['id']; ?>" value="accept"/>

Here you do not echo $rows1['id'], so if you have a look in the generated code, the name should be empty.

Correct this to

<input type="submit" name="<?php echo $rows1['id']; ?>" value="accept"/>

---

Furthermore in leaveactions.php you have the following code:

// ...
echo $_POST["$ii"];
if(isset($_POST['$ii'])){
  echo "accepted "; echo $_POST['$ii'];
// ...

Here you should first check, whether the variable (here it is $_POST[$ii] – no " needed) before doing the output.

As a result of the previous error, $_POST[$ii] is not set, thus you get the notice on the first echo and never enter the if-clause afterward.