Question

我在PHP和SQL中创建了一个简单的博客脚本，在需要显示最新帖子的页面上一切正常，但是，我创建了一个页面'comments.php'，它将加载所选帖子的评论。如果你点击帖子1，那么PHP生成的链接就是'comments.php？id = 1'。

这是我在'blog.php'页面上的代码，其中一切正常，发布内容正确显示：

 $db = new PDO('mysql:hostname=localhost;dbname=test;charset=utf8', 'root', '');
$req = $db->query('SELECT id, title, content, DATE_FORMAT(creation, \'%d/%m/%Y at %H:%i\') AS datePosted, user FROM blogpost ORDER BY creation DESC LIMIT 0, 5');
    while($data = $req->fetch()){
      ?>
        <div class="blogpost-container">
          <div class="blogpost-title">

            <?php
                echo htmlspecialchars($data['title']) . " | <em>Posted on ".$data['datePosted']."</em>";
             ?>
          </div>
          <div class="blogpost-content">
            <?php
                echo nl2br(htmlspecialchars($data['content']));
            ?>
          </div>
          <div class="blogpost-user">
            <?php
                echo "Posted by: ".$data['user'].".";
             ?>
           </div>
           <div class="blogpost-comments">
             <a href=<?php echo "'comments.php?id=". $data['id']."'"; ?> >
               Show comments
             </a>
           </div>
        </div>
      <?php
    }
    $req->closeCursor();
 ?>

这里，post post正确加载，但是一旦我继续使用comments.php？id = 1，它就没有显示任何内容。这是'comments.php'中的代码：

<?php
  $db = new PDO('mysql:hostname=localhost;dbname=test;charset=utf8;', 'root', '');
  $req = $db->prepare('SELECT id, title, content, DATE_FORMAT(creation, \'%d/%m/%Y at %H:%i), user FROM blogpost WHERE id = ?');
  $req->execute(array(
      $_GET['id']
  ));

  $data = $req->fetch();
 ?>
 <div class="blogpost-container">
   <div class="blogpost-title">

     <?php
         echo htmlspecialchars($data['title']) . " | <em>Posted on ".$data['datePosted']."</em>";
      ?>
   </div>
   <div class="blogpost-content">
     <?php
         echo nl2br(htmlspecialchars($data['content']));
     ?>
   </div>
   <div class="blogpost-user">
     <?php
         echo "Posted by: ".$data['user'].".";
      ?>
    </div>
 </div>

Answer 1

好的，我找到了我犯了错误的地方。我忘记添加\＆＃39;在

for /f "tokens=1 delims=." %%g in ('dir /b C:\Users\oRookie\Desktop\batch') do echo %%g >> out.txt

服务器未返回SQL数据

1 个答案: