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php - 如何获取表中特定记录的ID - PHP

问题描述

我正在尝试使用切换开关更改 MySQL 文章表中 is_public 列的值。如果我将输入（切换）包装在内部以使用 $_GET 方法在articles.php中进行必要的更改来获取记录的ID，那么它不会将is_public的数据获取到切换中。有没有其他方法可以获取特定记录的 id？

文章列表截图

这是 all-articles.php 代码：

                    <table id="datatable" class="table table-hover" style="width:100%">                                                                    
                         <thead>
                        <tr>
                            <th>#</th>
                            <th>Title</th>
                            <th>Author</th>
                            <th>Abstract</th>
                            <th>category</th>
                            <th>Date</th>
                            <th>Action</th> 
                            <th>Publish</th> 
                        </tr>
                    </thead>
                    <tbody>

                       <?php foreach($articles as $key => $article): ?>

                        <tr>
                            <td><?php echo $key+1 ?></td> 
                            <td><?php echo $article['title'] ?></td>
                            <td><?php echo $article['author_id'] ?></td>
                            <td><?php echo $article['abstract'] ?></td> 
                            <td><?php echo $article['category_id'] ?></td>
                            <td><?php echo $article['entry_date'] ?></td>
                            <td>
                              <div class="btn-group" role="group" aria-label="Basic example">
                                <button type="button" class="btn btn-primary btn-sm">Edit</button>
                                <button type="button" class="btn btn-success btn-sm">Show</button>
                                <button type="button" class="btn btn-danger btn-sm">Delete</button> 
                              </div>
                            </td>
                            <td> 

                                <form action="all-articles.php" method="post">
                                <label class="custom-toggle">
                                  <?php if($article['is_public']): ?>
                                    <input type="checkbox" checked name="check">
                                  <?php else: ?>
                                    <input type="checkbox"  name="uncheck">
                                  <?php endif; ?>
                                    <span class="custom-toggle-slider rounded-circle"></span>
                                </label> 
                                </form> 
                            </td>
                        </tr> 


                       <?php endforeach; ?>

                    </tbody>
                </table>

标签： phpmysqltoggle

解决方案

好吧，您可以在表单中发送文章 ID，然后收听它并用它做您想做的事情。所以，

// As you are already in foreach loop so,
 <form action="all-articles.php" method="post">
    <label class="custom-toggle">
      <?php if($article['is_public']): ?>
        <input type="checkbox" checked name="published" value="<?php echo $article['id']; ?>>
       <?php else: ?>
         <input type="checkbox"  name="published" value="<?php echo $article['id']; ?>>
       <?php endif; ?>
         <span class="custom-toggle-slider rounded-circle"></span>
      </label> 
  </form>

然后在你/all-articles.php检查$_POST['check']which 应该给article id. 使用该 ID 查询您的数据库并切换is_public列

另一种方法是发送一个隐藏字段，例如

// As you are already in foreach loop so,
 <form action="all-articles.php" method="post">
    <label class="custom-toggle">
      <?php if($article['is_public']): ?>
        <input type="checkbox" checked name="published" value="<?php echo $article['id']; ?>>
        <input type="hidden" name="is_published" value="yes"> or 1
       <?php else: ?>
         <input type="hidden" name="is_published" value="no"> or 0
         <input type="checkbox"  name="published" value="<?php echo $article['id']; ?>>
       <?php endif; ?>
         <span class="custom-toggle-slider rounded-circle"></span>
      </label> 
  </form>

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