Question

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

＆＃39;作者＆＃39;数据在作者字段中正确显示

＆＃39;关键字＆＃39;数据显示正确

然而，来自＆＃39;内容＆＃39;的数据变量不以textarea的形式显示。我知道声明中缺少值=，但这没有什么区别

我使用echo来确认$ content变量包含所需的数据

我错过了一些明显的东西吗？任何帮助非常感谢：）

<?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

Answer 1

您正在错误地访问$content

$content = $row_content['post_content'];

当您将查询结果存储在$row_posts时，它应该是

$content = $row_posts['post_content'];

Answer 2

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'];
      }
  }
 ?>

Answer 3

使用此$content = $row_posts['post_content'];

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

php表单echo不工作

3 个答案: