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我的页面上有一个我无法识别的错误，当我在线查看该网站时，它会将我的所有网站内容替换为值“0”。为什么我看不到表格，它是否与 SQL 服务器正常通信？

PS：登录工作并创建会话，因此 connect.php 确实可以正常工作。

这是members.php的代码：

<?php 
// starting session
session_start();

// check if user is logged in
if (!isset($_SESSION['username']))
{
    header('Location: http://wwww.gjertgjersund.com/');
    exit();
}
else
{
    // database connection
    require ('connect.php');

    //post record count
    $post_count = mysql_query("SELECT * FROM posts");
    $post_count_result = mysql_num_rows($post_count);

    //comment count
    $comment_count = mysql_query("SELECT * FROM comments");
    $comment_count_result = mysql_num_rows($comment_count);

    if(isset($_POST['submit']))
    {
        $newcategory = $_POST['newcategory'];
        if(!empty($newcategory))
        {
            $query = mysql_query("INSERT * INTO categories (category) VALUES ('$newcategory')";
            if($query)
        {
            echo 'New category added';
        }
        }
        else
        {
            echo 'Error';
        }
    }
    else
    {
        echo 'Missing newcategory';
    }
}
?>

<!DOCTYPE HTML>
<html>
<head>
<title> Folder </title>
<body>

<div class="wrap">
    <div id="menu">
        <ul>
            <li><a href='#'>Home</a></li>
            <li><a href='#'>Create New Post</a></li>
            <li><a href='#'>Delete Post</a></li>
            <li><a href='logout.php'>Log Out</a></li>
            <li><a href='#'>Blog Home Page</a></li>
        </ul>
    </div>
    <div id="maincontent">
        <table>
            <tr>
                <td>Total Blog Post</td>
                <td><?php echo $post_count_result ?></td>
            </tr>
            <tr>
                <td>Total Comments</td>
                <td><?php echo $comment_count_result ?></td>
            </tr>
        </table>
        <div id="categoryform">
            <form action="members.php" method="post">
                <label for="category">Add New Category</label>
                <input type="text" name="newcategory"/>
                <input type="submit" name="submit" value="Create"/>
            </form>
        </div>
    </div>
</div>

</body>
</html>
4

1 回答 1

0 
$query = mysql_query("INSERT * INTO categories (category) VALUES ('$newcategory')";

应该

$query = mysql_query("INSERT INTO categories (category) VALUES ('$newcategory')";
于 2013-06-19T05:39:24.537 回答