【问题标题】:Sweetalert won't show but it will updateSweetalert 不会显示,但会更新
【发布时间】:2021-09-26 08:26:41
【问题描述】:

我想在单击设置按钮后显示一个sweetalert,但它不起作用。这是我的索引页面,设置按钮可以工作,但不会显示甜蜜警报。可能是什么问题,我该怎么办?

index.php

 <form method='post' action='updataStatus.php'>
                <button type='submit' name='but_update' class="inline-block float ml-2 mt-1 btn-group pull-right btn-danger btn-sm">SET</button><button type="submit" id="dataExport" name="dataExport" value="Export to excel" class="inline-block float ml-2 mt-1 btn-group pull-right btn-info btn-sm">Export</button>
                <div class="table-responsive">

                <br>
                        

            

                  <tbody><table class="table table-hover table-bordered" id="sampleTable2">
                    <thead>
                      <tr>
                      <th><input type="checkbox" class="select-all checkbox" name="select-all" id="checkAll" /></th>
                      <th>Name</th>
                      <th>Scholarship Program</th>
                      <th>Course</th>
                      <th>Semester</th>
                      <th>Allowance</th>
                    </tr>
                  </thead>
                   <?php 
                    require_once "connection.php";
                    $query = "SELECT * FROM allowance";
                    $result = mysqli_query($conn,$query);

                    while($row = mysqli_fetch_array($result) ){
                        $id = $row['id'];
                        $Name = $row['Name'];
                        $Scholarship = $row['Scholarship'];
                        $Course = $row['Course'];
                        $Semester = $row['Semester'];
                        $statusAllowance = $row['statusAllowance'];
                    ?>
                        <tr>

                            <!-- Checkbox -->
                            <td><input type='checkbox' name='update[]' value='<?= $id ?>' ></td>
                            <td><p name="Name"><?php echo $row['Name']; ?></p></td>
                            <td><p name="Scholarship"><?php echo $row['Scholarship'] ?></p></td>
                            <td><p name="Course"><?php echo $row['Course'] ?></p></td>
                            <td><p name="Semester"><?php echo $row['Semester'] ?></p></td>
                            <td><p name='statusAllowance_<?= $id ?>'><?php echo $row['statusAllowance'] ?></td>

                        </tr>
                    <

?php
    
                        }
                        ?>
                      </table>
                    </tbody>

 <?php
      if(isset($_SESSION['success']) && $_SESSION['success'] !='')
      {
        ?>
        <script type="text/javascript">
          swal({
            title: "<?php echo $_SESSION['success']; ?>",
            icon: "<?php echo $_SESSION['status_code']; ?>",
            button: "yissh",
          });

        </script>

        <?php
          unset($_SESSION['success']);
      }

      ?>

这是我在编辑部分的代码,它有效,只有警报不会出现。 更新数据状态.php

 <?php
            require_once "connection.php";

              if(isset($_POST['but_update'])){

                  if(isset($_POST['update'])){
                      foreach($_POST['update'] as $id){
                          $statusAllowance = 'Received';

                                    if($statusAllowance != '' ){
                                        $updateUser = "UPDATE allowance SET statusAllowance='".$statusAllowance."' WHERE id=".$id;
                                        $query_run = mysqli_query($conn,$updateUser);

                                        if($query_run){
                                          $_SESSION['success'] = "YOUR DATA UPDATED";
                                          header('Location: tracking.php');
                                        }else{
                                          $_SESSION['success'] = "YOUR DATA IS NOT UPDATED";
                                          header('Location: tracking.php');
                                        }

                                    
                                }
                            }
                            
                        }
                      }

                ?>

【问题讨论】:

    标签: javascript php forms session sweetalert


    【解决方案1】:

    考虑到你还没有实现另一个函数来调用sweetalert;默认情况下,它应该是Swal.fire({}) 而不仅仅是swal({})

    https://sweetalert2.github.io/

    【讨论】:

    • 然后在 sweetalerts 网站上复制任何警报代码并在开发工具上复制粘贴,然后按 Enter。如果触发,则需要更多代码(例如必须检查是否包含 session_start()),但如果没有,则表示您尚未正确初始化 sweetalert(尝试在 &lt;head&gt;&lt;/head&gt; 中包含 cdn 脚本标签)
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