Question

我正在尝试使用PHP PDO构建一个简单的博客，但我在验证方面有点困惑/如果其他因为过去曾经发生在无类别的凌乱版本中它会说＆＃34 ;这篇文章不存在＆＃34;但是现在它只显示带有空框的页面，所以我想知道如何将if / else语句添加到类中以使其工作并在id不是数据库中匹配的时显示消息

 public function fetch_data($pid){

    try{
    global $pdo;

    $query = $pdo->prepare('SELECT * FROM post where post_id = ? order by post_date desc');
    $query->BindValue(1,$pid);
    $query->execute();

    return $query->fetch();

    }
     catch(PDOException $e) {
      echo '{"error":{"text":'. $e->getMessage() .'}}'; 
      }
     }

这是代码的公共函数位，而article.php页面代码是：

<?php

include_once('functions/main.php'); 
$post  = new Main;
$check = new Main;
$check_login = $check->logged_in();

if(isset($_GET['id'])){
    $pid = $_GET['id'];
    $post = $post->fetch_data($pid);

    $query = $pdo->prepare("UPDATE post SET post_views = post_views + 1 WHERE post_id = ?");
$query->execute(array($pid));
    ?>
<html>
    <head>
        <title><?php echo $post['post_title'];?></title>
          <meta name="viewport" content="width=device-width, initial-scale=1">


      <style>
.customimage{
background: url('<?php echo $post['post_image'];?>') !important;
}
</style>


    </head>


<body>

          <div class="pusher">
    <!-- Site content !-->
<div class="ui inverted vertical masthead center aligned segment purple customimage">
 <div class="ui text">
      <h1 class="ui inverted header">
        <?php echo $post['post_title'];?></h1>
              <br>
              <div class="ui black inverted label"> <i class="calendar icon"></i><?php echo $post['post_date'];?></div><div class="ui black inverted label"><i class="user icon"></i> <?php echo $post['post_author'];?></div><div class="ui black inverted label"><i class="unhide icon"></i> <?php echo $post['post_views']?></div>

  </div>
</div>

<div class="ui divider hidden"></div>

<div class="ui container">
<div class="ui segments">
  <div class="ui segment purple">
  <h1 class="ui header">
  <div class="content">
    <?php echo $post['post_title'];?>
     </div>
</h1>
  </div>
    <div class="ui segment">
    <?php echo $post['post_content'];?>
  </div>
  <div class="ui secondary segment">
    <button class="ui labeled icon button">
  <i class="left arrow icon"></i>
  Return to Posts</button>
  </div>
</div>
</div>

</div>
    </body>
</html>

    <?php
}else{
    header('Location:index.php');
}

?>

我无法弄清楚如何制作它，以便当你去？id = 876799然后说文章不存在但目前它只是空白。

Answer 1

在查询后和显示结果时检查$post的值。

$post = $post->fetch_data($pid);
if ($post) {
    $query = $pdo->prepare("UPDATE post SET post_views = post_views + 1 WHERE post_id = ?");
    $query->execute(array($pid));
} else {
    display_post_not_found($pid);
    exit();
}
?>
<html>
    ...
</html>

在display_post_not_found()函数（您必须编写）中，您可以显示有关错误的信息页面，或者只是重定向到某个地方。

完整代码：

<?php

include_once('functions/main.php'); 
$main  = new Main;
$check = new Main;
$check_login = $check->logged_in();

if(isset($_GET['id'])){
    $pid = $_GET['id'];
    $post = $main->fetch_data($pid);
    if ($post) {
        $query = $pdo->prepare("UPDATE post SET post_views = post_views + 1 WHERE post_id = ?");
        $query->execute(array($pid));
    } else {
        display_post_not_found($pid);
        exit();
    }
    ?>
<html>
    <head>
        <title><?php echo $post['post_title'];?></title>
          <meta name="viewport" content="width=device-width, initial-scale=1">


      <style>
.customimage{
background: url('<?php echo $post['post_image'];?>') !important;
}
</style>


    </head>


<body>

          <div class="pusher">
    <!-- Site content !-->
<div class="ui inverted vertical masthead center aligned segment purple customimage">
 <div class="ui text">
      <h1 class="ui inverted header">
        <?php echo $post['post_title'];?></h1>
              <br>
              <div class="ui black inverted label"> <i class="calendar icon"></i><?php echo $post['post_date'];?></div><div class="ui black inverted label"><i class="user icon"></i> <?php echo $post['post_author'];?></div><div class="ui black inverted label"><i class="unhide icon"></i> <?php echo $post['post_views']?></div>

  </div>
</div>

<div class="ui divider hidden"></div>

<div class="ui container">
<div class="ui segments">
  <div class="ui segment purple">
  <h1 class="ui header">
  <div class="content">
    <?php echo $post['post_title'];?>
     </div>
</h1>
  </div>
    <div class="ui segment">
    <?php echo $post['post_content'];?>
  </div>
  <div class="ui secondary segment">
    <button class="ui labeled icon button">
  <i class="left arrow icon"></i>
  Return to Posts</button>
  </div>
</div>
</div>

</div>
    </body>
</html>

    <?php
}else{
    header('Location:index.php');
}

function display_post_not_found($pid) {
    echo "Post $pid could not be found";
}

Answer 2

你有两个选择：

重定向到显示错误的页面，例如post_not_found.php或其他内容。
使用if / else语句包装您现在拥有的整个html页面，该语句将根据您的条件输出不同的内容。

我建议你选择第一个选项，你需要做的是检查$post是否有数据，如果它没有header("Location: post_not_found.php");

在PHP中向公共类函数添加validation / if else语句

2 个答案: