Question

我正试图让我的帖子更新，以防我第一次在我的网站上发布文章时出错。

不确定我在这里做错了什么。

这是我的更新代码：

    <div class="row">
        <?php
$post_title = "";
$description = "";
$id = $_GET['id'];

$result = mysql_query("SELECT title, description FROM htp_news WHERE id='$id'");

$post_title     = mysql_result($result,0,"title");
$description    = mysql_result($result,0,"description");

?>

    <div class="row">

<form method="post" action="update-news.php">
<input type="hidden" name="ud_id" style="width: 100%" value="<? echo "$id"; ?>">

<div class="grid_12 botspacer60">


Title: <input type="text" name="ud_title" value="<?php echo "$post_title"; ?>">
<br /><br />

News Details:<br />
<textarea id="tiny_mce" name="ud_description" rows="8"><?php echo "$description"; ?></textarea>

</div>


<div class="grid_12">

    <input type="submit" value="Update">
    <input type="button" value="Cancel" onclick="window.location = '/admin'">
    </div>
</form>
</div>


</div>

这是我的行动页面：

<?php
include($_SERVER['DOCUMENT_ROOT'] . "/includes/database.php");

$ud_id          = $_POST['ud_id'];
$ud_title       = $_POST['ud_title'];
$ud_description = $_POST['ud_description'];

// Insert record into database by executing the following query:
$query="UPDATE htp_news SET title='$ud_title', description='$ud_description' "."WHERE id='$ud_id'";

mysql_query($query);


echo "The post has been updated.<br />
<a href='edit-delete-news.php'>Update another position.</a><br />";

mysql_close();
?>

我很欣赏有关此事的任何指导。

Answer 1

在查询中WHERE Clause之前添加空格。

Answer 2

使用以下 -

$query="UPDATE htp_news SET title='$ud_title', description='$ud_description'  WHERE id='$ud_id'";

Answer 3

试试这个，你需要在查询中引用

$result = mysql_query("SELECT `title`, `description` FROM `htp_news` WHERE id='$id'");
$query="UPDATE htp_news SET `title`='".$ud_title."', `description`='".$ud_description."' "." WHERE `id`='".$ud_id."'";

从数据库更新帖子

3 个答案: