Cannot delete row in database using PHP

Question

I can't seem to delete row in database by id in php I think the the id is not passed to the $_POST['delete'] however, the popup "Your data is deleted" is displayed, but the data is not deleted. So I'm not sure where is the error in this code.

I also try to delete the data by its id for example: Delete book where no='4'; and the code seems to run fine because the data is deleted in the database.

<html>

<script>
  function confirmDelete() {
    return confirm('Are you sure?');
  }
</script>


<!DOCTYPE html>

<head>
  <form action="test.php" method="POST">

    <br><br><br>

    <table bordercolor="#FFCC66" align="center" bgcolor="#FFFFFF">
      <tr>
        <th>No</th>
        <th>Title</th>
        <th>Author</th>
        <th>Year</th>
        <th>Donor's Name</th>
        <th>Call Number</th>
        <th>Date Received</th>
        <th>Handled By</th>
        <th></th>
        <th></th>
      </tr>

      <?php
  include ('config.php');

  $view=mysqli_query($conn,"SELECT * FROM book");
  
  
?>
        <?php while($v=mysqli_fetch_array($view)){ ?>
        <tr>

          <td>
            <?php echo $v["no"];?>
          </td>
          <td>
            <?php echo $v["title"];?>
          </td>
          <td>
            <?php echo $v["author"];?>
          </td>
          <td>
            <?php echo $v["year"];?>
          </td>
          <td>
            <?php echo $v["donorname"];?>
          </td>
          <td>
            <?php echo $v["callnum"];?>
          </td>
          <td>
            <?php echo $v["datereceived"];?>
          </td>
          <td>
            <?php echo $v["handledby"];?>
          </td>
          <td><input type="submit" name="delete" value="Delete" onclick="return confirmDelete('Are you sure?');" /></td>
        </tr>
        <?php
    } ?>

          </tr>
    </table>
    <br><br>

  </form>
  </body>

</html>

<?php

if(isset($_POST['delete']))
{
    
    include('config.php');
    
  $no =$v["no"];
  $d=mysqli_query($conn,"DELETE FROM `book` WHERE no='$no'");
  if ($d)
  {

    echo "<script type='text/javascript'> alert('Your data is deleted!'); </script>";
    echo"<meta http-equiv='Refresh' content='0' >"; 
  }
  else
  {
    echo "<script type='text/javascript'> alert('Your data cannot delete!'); </script>";
  }
     mysqli_close($conn);

}
  
?>

Rakesh Jakhar · Accepted Answer · 2019-09-16 06:01:28Z

1

Change the submit element to

<td>
  <input type="submit" name="delete" value="<?php echo $v['no'];?>" onclick="return confirmDelete('Are you sure?');" />
</td>

and

$no = $_POST["delete"];

answered Sep 16, 2019 at 6:01

Rakesh Jakhar

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Comments

Meloman · Accepted Answer · 2019-09-16 06:19:00Z

0

Another solution si to add a hidden input with your value.

<td>
    <?php echo $v["no"];?>
    <input type="hidden" value="<?php echo $v["no"];?>" />
</td>

In your php you will find the value in $_POST['no']

This solution is better to pass multiple arguments in POST like a captcha or a confirmation (checkbox).

answered Sep 16, 2019 at 6:19

Meloman

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Comments

Sanjun Dev · Accepted Answer · 2019-09-16 06:35:37Z

0

logic is not correct, while you press the delete button, all the data will be passed along with submitting because your tag is outside of the loop.

As my opinion, you should use ajax like functionality here, or follow this method.

<?php while($v=mysqli_fetch_array($view)){ ?>

<form action="test.php" method="POST">
   <tr>

      <td>
        <?php echo $v["no"];?>
        <input type="hidden" value="<?php echo $v["no"];?>" name="no" >
      </td>
      <td><input type="submit" name="delete" value="Delete" onclick="return confirmDelete('Are you sure?');" /></td>
    </tr>
</form>
<?php } ?>

and in your post call use $no = $_POST['no']; instead of $no =$v["no"];

answered Sep 16, 2019 at 6:35

Sanjun Dev

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