php表单回显不工作

我正在运行select语句来检索db值。然后，将检索到的数据显示/回显到表单字段中

“作者”数据在“作者”字段中正确显示

“关键字”数据显示正确

但是，来自“content”变量的数据没有显示在表单文本区域中。 我知道语句中缺少值=，但这没有什么区别

我已经使用echo确认$content变量包含所需的数据

我错过了什么明显的东西吗？非常感谢您的帮助：

<?php

  if(isset($_GET['edit_post'])) { 
  $edit_id = $_GET['edit_post'];

  $select_post = "select * FROM posts WHERE post_id = '$edit_id'";  
  run_query = mysql_query($select_post);
  while($row_posts = mysql_fetch_array($run_query)) {

  $author = $row_posts['post_author'];
  $keywords = $row_posts['post_keywords'];     
  $content = $row_content['post_content'];
      }
  }
 ?>




      <tr> 
        <td align="right" bgcolor="#ccc"> Author:</td>
        <td><input type="text" name="post_author" value ="<?php echo $author;?>"/></td>
    </tr>   



     <tr> 
        <td align="right" bgcolor="#ccc"> Keywords:</td>
        <td><input type="text" name="post_keywords" size="100" value ="<?php echo     $keywords;?>"/></td>
    </tr>   



    <tr> 
        <td align="right" bgcolor="#ccc"> Content:</td>
        <td><textarea name="post_content" rows="10" cols="60"><?php echo $content;?>             </textarea></td>
    </tr>       

非常感谢,， P

您错误地访问了$content

当您将查询结果存储在$row_posts中时，它应该是

$content = $row_posts['post_content'];

您错误地访问了$content

当您将查询结果存储在$row_posts中时，它应该是

$content = $row_posts['post_content'];

试试这个

<?php

  if(isset($_GET['edit_post'])) { 
  $edit_id = $_GET['edit_post'];

  $select_post = "select * FROM posts WHERE post_id = '$edit_id'";  
  $run_query = mysql_query($select_post); 
  $author = "";
  $keywords = "";     
  $content = "";
  if(mysql_num_rows($run_query)>0) {
  $row_posts = mysql_fetch_array($run_query);
  $author = $row_posts['post_author'];
  $keywords = $row_posts['post_keywords'];     
  $content = $row_posts['post_content'];
      }
  }
 ?>

试试这个

<?php

  if(isset($_GET['edit_post'])) { 
  $edit_id = $_GET['edit_post'];

  $select_post = "select * FROM posts WHERE post_id = '$edit_id'";  
  $run_query = mysql_query($select_post); 
  $author = "";
  $keywords = "";     
  $content = "";
  if(mysql_num_rows($run_query)>0) {
  $row_posts = mysql_fetch_array($run_query);
  $author = $row_posts['post_author'];
  $keywords = $row_posts['post_keywords'];     
  $content = $row_posts['post_content'];
      }
  }
 ?>

使用这个$content=$row_posts['post_content']

而不是$content=$row_content['post_content']

使用这个$content=$row_posts['post_content']

而不是$content=$row_content['post_content']