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当我单击删除按钮时,它只是刷新而不是删除

[英]When I click delete button it's just refreshing but not deleting

When I click delete button in popup modal it's just refreshing but not deleting, I think it's bad URL, what can I do?当我在弹出模式中单击删除按钮时,它只是刷新而不是删除,我认为它很糟糕 URL,我该怎么办? I added all my code, Please help me我添加了所有代码,请帮助我

<form method="post">
        <input type="hidden" name="delete" value="<?php echo $darb_id ?>">
     <?php
    echo "<td><a rel={$darb_id} href='javascript:void(0);' class='btn btn-danger delete_link'>Ištrinti</a></td>";?>

    </form>

MY DELETE MODAL CODE:我的删除模式代码:

<!-- Modal -->
<div id="myModal" class="modal fade" role="dialog">
  <div class="modal-dialog">

    <!-- Modal content-->
    <div class="modal-content">
      <div class="modal-header">
        <button type="button" class="close" data-dismiss="modal">×</button>
        <h4 class="modal-title">DELETE message</h4>
      </div>
      <div class="modal-body">
        <h3 class="text-center"></h3>
      </div>
      <div class="modal-footer">
        <a href="" class="btn btn-danger modal_delete_link">DELETE</a>
        <button type="button" class="btn btn-default" data-dismiss="modal">CANCEL</button>
      </div>
    </div>

  </div>
</div>

My JAVASCRIPT:我的 JAVASCRIPT:

$(document).ready(function(){

      $(".delete_link").on('click', function(){

          var id = $(this).attr("rel");

          var delete_url = "darb.php?delete="+ id;

          $(".modal_delete_link").attr("href", delete_url);

          $("#myModal").modal('show');

      });
  });

My darb.php code where everything happens我的 darb.php 代码,一切都发生在这里

<?php

include("delete_modal.php");
include("look_darb.php");


    $query = "SELECT * FROM darb ORDER BY darb_id DESC ";
    $select_darb = mysqli_query($connection, $query);

    while ($row = mysqli_fetch_assoc($select_darb)) {
        $darb_id            = $row['darb_id'];
        $darb_user          = $row['darb_user'];
        $darb_title         = $row['darb_title'];
        $darb_status        = $row['darb_status'];
        $darb_date          = $row['darb_date'];
        $darb_work          = $row['darb_work'];
        $darb_finish        = $row['darb_finish'];
        $darb_result        = $row['darb_result'];
        $darb_fileUpload    = $row['darb_fileUpload'];
        $darb_payment       = $row['darb_payment'];

        echo "<tr>"; ?>

 <td><input class='checkBoxes' type='checkbox' name='checkBoxArray[]' value='<?php echo $darb_id; ?>'></td>

        <?php

        echo "<td>$darb_id </td>";

        echo "<td>$darb_user</td>";

        echo "<td>$darb_title</td>";

        echo "<td>$darb_status</td>";
        echo "<td>$darb_date</td>";

        echo "<td>$darb_result</td>";

        echo "<td><a href='../uploads/$darb_fileUpload'>$darb_fileUpload</a></td>";

        echo "<td>$darb_payment</td>";
        ?>
        <form method="post">
            <input type="hidden" name="darb_id" value="<?php echo $darb_id ?>">
         <?php
          echo "<td><a rel={$darb_id} href='javascript:void(0);' class='btn btn-primary look_link'>Peržiūrėti</a></td>";?>
        </form>
        <?php

        echo "<td><a class='btn btn-info' href='darb.php?source=edit_darb&d_id={$darb_id}'>Redaguoti</a></td>";?>
        <form method="post">
            <input type="hidden" name="delete" value="<?php echo $darb_id ?>">
         <?php
        echo "<td><a rel={$darb_id} href='javascript:void(0);' class='btn btn-danger delete_link'>Ištrinti</a></td>";?>

        </form>
        <?php
    }

      ?>
          </tbody>
        </table>
      </form>

<?php

if (isset($_POST['delete'])) {
    $the_darb_id = escape($_POST['darb_id']);

    $query = "DELETE FROM darb WHERE darb_id = {$the_darb_id} ";
    $delete_query = mysqli_query($connection, $query);
    header("Location: /pvd/admin/darb.php");
}



?>

I have tried a lot of advice online but couldn't resolve this problem.我在网上尝试了很多建议,但无法解决这个问题。

Change this:改变这个:

$the_darb_id = escape($_POST['darb_id']);

to this:对此:

$the_darb_id = escape($_POST['delete']);

Because there's no $_POST['darb_id'] .因为没有$_POST['darb_id']

PS You also might want to DELETE before you SELECT . PS您可能还想在DELETE之前SELECT

Full code solved:完整代码已解决:

    <?php

include("delete_modal.php");
include("look_darb.php");


    $query = "SELECT * FROM darb ORDER BY darb_id DESC ";
    $select_darb = mysqli_query($connection, $query);

    while ($row = mysqli_fetch_assoc($select_darb)) {
        $darb_id            = $row['darb_id'];
        $darb_user          = $row['darb_user'];
        $darb_title         = $row['darb_title'];
        $darb_status        = $row['darb_status'];
        $darb_date          = $row['darb_date'];
        $darb_work          = $row['darb_work'];
        $darb_finish        = $row['darb_finish'];
        $darb_result        = $row['darb_result'];
        $darb_fileUpload    = $row['darb_fileUpload'];
        $darb_payment       = $row['darb_payment'];

        echo "<tr>"; ?>

 <td><input class='checkBoxes' type='checkbox' name='checkBoxArray[]' value='<?php echo $darb_id; ?>'></td>

        <?php

        echo "<td>$darb_id </td>";

        echo "<td>$darb_user</td>";

        echo "<td>$darb_title</td>";

        echo "<td>$darb_status</td>";
        echo "<td>$darb_date</td>";

        echo "<td>$darb_result</td>";

        echo "<td><a href='../uploads/$darb_fileUpload'>$darb_fileUpload</a></td>";

        echo "<td>$darb_payment</td>";
        ?>
        <form method="post">
            <input type="hidden" name="darb_id" value="<?php echo $darb_id ?>">
         <?php
          echo "<td><a rel={$darb_id} href='javascript:void(0);' class='btn btn-primary look_link'>Peržiūrėti</a></td>";?>
        </form>
        <?php

        echo "<td><a class='btn btn-info' href='darb.php?source=edit_darb&d_id={$darb_id}'>Redaguoti</a></td>";?>
        <form method="post">
            <input type="hidden" name="delete" value="<?php echo $darb_id ?>">
         <?php
        echo "<td><a rel={$darb_id} href='javascript:void(0);' class='btn btn-danger delete_link'>Ištrinti</a></td>";?>

        </form>
        <?php
    }

      ?>
          </tbody>
        </table>
      </form>

<?php

if (isset($_POST['delete'])) {
    $the_darb_id = escape($_POST['delete']);

    $query = "DELETE FROM darb WHERE darb_id = {$the_darb_id} ";
    $delete_query = mysqli_query($connection, $query);
    header("Location: /pvd/admin/darb.php");
}



?>

Explanation of the solution:解决方案说明:

if (isset($_POST['delete'])) {
//assign the variable  $the_darb_id to $_POST['delete'
    $the_darb_id = escape($_POST['delete']);

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