Question

提前致谢。

我正在尝试更新我的数据库中的信息，但我没有运气，也没有错误消息。

这是页面代码（pid或post_id在URL中）：

<?php
include('page with functions.php');


if (isset($_GET['pid'], $_POST['title'], $_POST['body'])){

if (edit_post($_GET['pid'], $_POST['title'], $_POST['body'])){
header("Location: blog_edit.php?pid={$_GET['pid']}");
}else{
header("Location: some location.php");
}

die();

}

?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"     "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Blog Edit</title>
</head>

<body>

<div>
    <?php

if (isset($_GET['pid']) ===false || valid_pid($_GET['pid']) === false){
echo 'Invalid post ID';
}else{
$post = get_post($_GET['pid']);

    ?>

<h2><?php echo $post['title']; ?></h2>


<hr />

<p><?php echo $post['body']; ?></p>

<hr />



    <form action="" method="post">
         <p>
<label for="title">Title</label>
        <input type="text" name="title" id="title" value="<?php echo $post['title']; ?>"/>
        </p>
        <p>
            <textarea name="body" rows="20" cols="60"><?php echo $post['body']; ?></textarea>
        </p>
        <p>
            <input type="submit" value="Edit Post" />
        </p>
    </form>
<?php
}
?>
</div>
</body>
</html>

这是函数（pid或post_id在URL中）：

// edits a blog entry
function edit_post($title, $body){

$title = mysql_real_escape_string(htmlentities($title));
$body = mysql_real_escape_string(nl2br(htmlentities($body)));

mysql_query("UPDATE `posts` (`post_title`, `post_body`) SET `post_title` = '{$title}' AND `post_body` = '{body}' WHERE `post_id` = {$pid}");

}

任何帮助都将非常感谢！仅供参考我是PHP MYSQL的新手，所以请善待。

Answer 1

if (edit_post($_GET['pid'], $_POST['title'], $_POST['body'])){
header("Location: blog_edit.php?pid={$_GET['pid']}");
}else{
header("Location: some location.php");
}

您在函数调用中传递了三个参数，但函数定义只有两个参数。

function edit_post($title, $body){

$title = mysql_real_escape_string(htmlentities($title));
$body = mysql_real_escape_string(nl2br(htmlentities($body)));

mysql_query("UPDATE `posts` (`post_title`, `post_body`) SET `post_title` = '{$title}' AND `post_body` = '{body}' WHERE `post_id` = {$pid}");

}

此外，从函数返回一个布尔值。

function edit_post($pid, $title, $body){

$title = mysql_real_escape_string(htmlentities($title));
$body = mysql_real_escape_string(nl2br(htmlentities($body)));

$result = mysql_query("UPDATE `posts` (`post_title`, `post_body`) SET `post_title` = '{$title}' AND `post_body` = '{body}' WHERE `post_id` = {$pid}");

echo $result;
}

MYSQL PHP通过html表单更新db表信息

1 个答案: