Question

我想通过ajax将用户ID从我的表传递到modal，但只想到我得到的是"undefined"。当我替换var dataString = 'id=' + recipient;时与var dataString = 'id=' + 1;它工作正常所以由于某种原因我无法获得点击变量的价值。我错过了什么？

PHP：

<?php

                        $query = mysql_query("select * from users");
                        $i=0;
                        while($fetch = mysql_fetch_array($query)):



                        echo '<tr>';
                        echo'<td> '.$fetch['user_id'].'</td>';
                        echo'<td> '.$fetch['user_name'].'</td>';
                        echo'<td> '.$fetch['user_email'].'</td>';
                        echo'<td> '.$fetch['user_imie'].'</td>';
                        echo'<td> '.$fetch['user_nazwisko'].'</td>';
                        echo'<td> '.$fetch['user_telefon'].'</td>';
                        echo'<td> '.$fetch['user_konto_akty'].'</td>';
                        echo'<td> '.$fetch['user_uprawnienia'].'</td>';
                        echo'<td>' .date("d.m.Y, H:i", $fetch['user_regdate']). '</td>';

                        echo'<td>  <a class="btn btn-primary btn-sm" href="Kasuj_tab.php?user_id='.$fetch['user_id'].'">Usuń</a></td>';

                       ?>
                                <td>
                                    <a class="btn btn-small btn-primary" data-toggle="modal" data-target="#exampleModal_user" data-whatever1="<?php echo $fetch['user_id']; ?>">Edit</a></td>

                    <?php
                            echo '</tr>';
                        endwhile;
                    ?>

的Ajax：

  $('#exampleModal_user').on('show.bs.modal', function (event) {
          var button = $(event.relatedTarget) // Button that triggered the modal
          var recipient = button.data('whatever1') // Extract info from data-* attributes
          var modal = $(this);
          var dataString = 'id=' + recipient;

            $.ajax({
                type: "GET",
                url: "Modal_user.php",
                data: dataString,
                cache: false,
                success: function (data) {
                    console.log(data);
                    modal.find('.ct').html(data);
                },
                error: function(err) {
                    console.log(err);
                }
            });  
    })

PHP：

<?php
include 'Panel_Logowanie/config.php';
db_connect();

$id = $_GET['id'];  
echo"$id";


?>

通过ajax将user_id从mysql传递到bootstrap模式

0 个答案: