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Php 分页我的博客不起作用_Php_Mysql_Pagination - Fatal编程技术网

Php 分页我的博客不起作用

Php 分页我的博客不起作用,php,mysql,pagination,Php,Mysql,Pagination,我正在尝试使用PHP、HTML和MySQL制作一个分页博客。我写的代码,但出于某种原因,网页显示为空白。我的代码怎么了?Chrome的控制台返回500内部服务器错误 <div id="article"> <?php include 'php/mysql_connect.php'; if(empty($_GET)){ $current_id = SELECT max(id) FROM posts; } else{ $current_id = my

我正在尝试使用PHP、HTML和MySQL制作一个分页博客。我写的代码,但出于某种原因,网页显示为空白。我的代码怎么了?Chrome的控制台返回500内部服务器错误

<div id="article">

<?php
  include 'php/mysql_connect.php';
  if(empty($_GET)){
    $current_id = SELECT max(id) FROM posts;
  }
  else{
    $current_id = mysql_safe_string($_GET['id']);
  }

  $result = mysql_safe_query('SELECT * FROM posts WHERE id=%s LIMIT 1',$current_id);
  if(!mysql_num_rows($result)){
    echo '<h2>No Posts Found</h2>';
    exit;
  }

  $row = mysql_fetch_assoc($result)
  echo '<h2>'.$row['title'].'</h2>';
  echo '<div class="row">';
  echo '  <div class="group1 col-sm-6 col-md-6">';
  echo '    <span class="glyphicon glyphicon-pencil"></span><a data-toggle="collapse" data-target="#comments" class"collapsed">'.$row['num_comments'].' Comments </a>';
  echo '    <span class="glyphicon glyphicon-time"></span>'.date('F j<\s\up>S</\s\up>, Y', $row['date']);
  echo '  </div>';
  echo '</div>';
  echo '<br />';
  echo '<p class="lead">'.n12br($row['body']).'</p>';
  ?> 

  <div id="comments" class="collapse" >
    <div class="well">
      <h4>Leave a comment</h4>
      <?php echo '<form role="form" method="post" action="php/comment_add.php?id=($current_id)" class="clearfix">'; ?>
        <div class="col-md-6 form-group">
          <label class="sr-only" for="name">Name</label>
          <input type="text" class="form-control" id="name" placeholder="Name" required />
        </div>
        <div class="col-md-6 form-group">
            <label class="sr-only" for="email">Email</label>
            <input type="email" class="form-control" id="email" placeholder="Email" required />
       </div>
        <div class="col-md-12 form-group">
            <label class="sr-only" for="content">Comment</label>
            <textarea class="form-control" id="content" placeholder="Comment" required></textarea>
        </div>
        <div class="col-md-12 form-group text-right">
            <button type="submit" class="btn btn-primary">Submit</button>
        </div>
      </form>
    </div>
    <br>

<?php 
  $result = mysql_safe_query('SELECT * FROM comments WHERE post_id=%s ORDER BY date ASC',$current_id);
  echo '  <ul id="comments" class="comments">';

  while($row = mysql_fetch_assoc($result)){
  echo '    <li class="comment">';
  echo '      <div id="inline" ><h4 style="display:inline;">'.$row['name'].'</h1><sup><p style="display:inline; font-size:10px;">&nbsp; '.date('j-M-Y g:ia', $row['date']).'</p></sup></div>';
  echo '      <em>'.n12br($row['content']).'</em>';
  echo '    </li>';
  echo '  </ul>';
  }
?>
    <hr>
  </div>
</div>

<nav>
  <ul class="pager">
<?php

    $newer_id = IFNULL(mysql_safe_query('SELECT min(id) FROM posts WHERE id > $current_id ORDER BY id ASC LIMIT 1'),-1);
    $older_id = IFNULL (mysql_safe_query('SELECT max(id) FROM posts WHERE id < $current_id ORDER BY id ASC LIMIT 1'),-1);

    if($newer_id != -1){
      echo '<li><a href="#">Newer</a></li>';
    }
    if ($older_id != -1){
      echo '<li><a href="#">Older</a></li>';
    }
?>
  </ul>
</nav>
如果您的文件是myfile.html,并且您没有嵌入式PHP的权限,那么它将不会在服务器上执行

if(empty($_GET)){
    $current_id = SELECT max(id) FROM posts;
}

sql周围没有引号是错误的。

正如other所指出的,您所面临的语法错误应该立即清楚,您的$current\u id查询没有被引用。一个良好的开端是修复第一个块:

if(empty($_GET)) {
    $current_id = "SELECT max(id) FROM posts;";
} else {
    $current_id = mysql_safe_string($_GET['id']);
}

如果你得到的是500,那么应该在某个地方有一个日志。发布它的相关部分将是明确您面临的错误的第一步。语法错误,第6行出现意外的“max”T_字符串ifempty$_GET{$current_id=SELECT maxid FROM posts}您使用的是旧教程。称为ext/mysql的mysql_*函数已弃用,不应使用。相反,您应该使用or。好吧,如果我删除所有mysql_safe_字符串和mysql_safe_查询,我将如何使用PDO使$current_id变量安全?啊,我明白了。非常感谢。它们是单引号还是双引号重要吗?
if(empty($_GET)){
    $current_id = SELECT max(id) FROM posts;
}
if(empty($_GET)) {
    $current_id = "SELECT max(id) FROM posts;";
} else {
    $current_id = mysql_safe_string($_GET['id']);
}