Question

我有四个php页面：login.php，db_connection.php，admin.php和blog.php

我希望能够通过admin.php页面动态地将文章插入blog.php。

我的代码是：

的login.php

<?php 
    error_reporting(E_ALL & ~E_NOTICE); 
    session_start();

    if($_POST['submit']) {
        $dbUserName = "admin";
        $dbPassword = "password";

        $username = strip_tags($_POST['username']);
        $username = strtolower($username);
        $password = strip_tags($_POST['password']);

        if ($username == $dbUserName && password == $dbPassword) {
            $_SESSION['username'] = $username;
            header('Location: admin.php');
        } else {
            echo "<h1 class='denied'>Access denied<h1>";
        }
    }
?>

<div id="login_box">
    <h1 class="loreg">Log in</h1>
    <form action='login.php' method='post' id='contact_form'>
        <input type='text' name='username' placeholder='username... *' id='email' maxlength='60'><br>
        <input type='password' name='password' placeholder='password... *' id='password' maxlength='30'><br>
        <input type='submit' class='button' value='log in' id='login'>
        <input type='reset' class='button' value='cancel' id='cancel'>
    </form>
</div>

db_connection

<?php  
    $dbCon = mysqli_connect("localhost", "root", "passw0rd", "worldtour") or die(mysqli_error());
?>

admin.php的

<?php 
    error_reporting(E_ALL & ~E_NOTICE); 
    session_start();

    if (isset($_SESSION['username'])) {
        $username = ucfirst($_SESSION['username']);

        if ($_POST['submit']) {
            $title = $_POST['title'];
            $submit = $_POST['subtitle'];
            $content = $_POST['content'];
            include_once("db_connection.php");
            $sql = "INSERT INTO blog (title, subtitle, content)
                    VALUE ('$title', '$subtitle', '$content')";
            mysqli_query($dbCon, $sql);
            echo "<h1>Blog entry posted</h1>";
        } 
    } else {
        header('Location: login.php');
        die();
    }
?>

<h1>Welcome, <?php echo $username; ?>!</h1>
<form action="admin.php" method="post">
    <h3>Title: </h3><input type="text" name="title"><br>
    <h3>Subtitle: </h3><input type="text" name="subtitle"><br>
    <h3>Content: </h3><textarea name="content" id="content" cols="30" rows="10"></textarea>
    <input type="submit" name="submit" value="post entry">
</form>
<br>
<a href="/sites/worldtour/public/blog.php">View blog entries</a> | <a href="/sites/worldtour/public/logout.php">Log out</a>

blog.php的

<?php 
    include_once("db_connection.php");

    $sql = "SELECT * FROM blog ORDER BY id DESC";
    $result = mysqli_query($dbCon, $sql);

    while($row = mysqli_fetch_array($result)) {
    $title = $row['title'];
    $subtitle = $row['subtitle'];
    $content = $row['content'];
?>
    <h1 class="headers"><?php echo $title; ?> <br> <small><?php echo $subtitle; ?></small></h1>
    <article><?php echo $content; ?></article>
    <hr class="artline">
<?php  
    }
?>

当我尝试通过login.php页面登录时，即使我插入正确的用户/密码，页面也会刷新，并且不会将我重定向到admin.php。当我插入错误的用户时，它不会提示我访问被拒绝的消息。我的代码中是否有任何错误或我遗漏的内容？感谢。

Answer 1

POST方法使用标记中的name属性来标识其值。在$ _POST ['submit']中，您实际上在寻找名称='submit'的标记。

提交按钮没有定义名称属性。您需要在

中添加name ='submit'

<input type='submit' class='button' value='log in' id='login'>

如何动态创建内容到数据库并在网页上显示？

1 个答案: