Question

我正在制作一个基本的歌曲推荐网站，我有一个表单设置，通向一个页面，上面有这个确切的代码：

<?php
        ini_set('display_errors',1);
        ob_start();
        session_start();

        $host = "localhost";
        $user = "root";
        $pass = "MYPASS";
        $db = "tts";

        $conn = mysqli_connect($host, $user, $pass, $db);

        $song = $_POST['song'];
        $artist = $_POST['artist'];
        $album = $_POST['album'];
        $linkitunes = $_POST['linkitunes'];
        $artwork = $_POST['artwork'];

        $song = stripslashes($song);
        $artist = stripslashes($artist);
        $album = stripslashes($album);
        $linkitunes = stripslashes($linkitunes);
        $artwork = stripslashes($artwork);

        $sql = "INSERT INTO recommendation (user_id, song, artist, album, linkitunes, artwork, rating)";
        $sql = $sql . "VALUES ($_SESSION['id'], '$song', '$artist', '$album', '$linkitunes', '$artwork', '$rating');";

        print "Hello.";

        $result = mysqli_query($sql) or die("Fail");

        ob_flush();
?>

它总是显示“你好”。字符串，直到我添加$ sql值。我认为代码语法有问题，但不确定。尝试了很多变化。为了以防万一，我也添加了表单代码：

<form action="recommend-action.php" method="POST">
    <div id="noP" align="center">
            <h2>Make a new Recommendation</h2>
            <p>Please <a href="song-search.php">search</a> for your song before you recommend it.</p>
    </div>
    <div align="center">
            <input required name="song" type="text" placeholder="Song" maxlength="50"></input>
            <input required name="artist" type="text" placeholder="Artist" maxlength="50"></input>
            <input name="album" type="text" placeholder="Album" maxlength="50"></input>
            <input name="artwork" type="url" placeholder="Artwork" maxlength="500"></input>
            <input name="linkitunes" type="url" placeholder="Link in iTunes" maxlength="500"></input>
            <input id="submit" type="submit" value="Recommend"></input>
    </div>

Answer 1

你应该执行这个......

        $result = mysqli_query($con,$sql); or die("Fail");

Answer 2

我对您的代码做了一些修改，试试这个：

<?php
ini_set('display_errors',1);
ob_start();
session_start();

$host = "localhost";
$user = "root";
$pass = "MYPASS";
$db = "tts";

$conn = mysqli_connect($host, $user, $pass, $db);

$song = $_POST['song'];
$artist = $_POST['artist'];
$album = $_POST['album'];
$linkitunes = $_POST['linkitunes'];
$artwork = $_POST['artwork'];

$song = stripslashes($song);
$artist = stripslashes($artist);
$album = stripslashes($album);
$linkitunes = stripslashes($linkitunes);
$artwork = stripslashes($artwork);

$sql = "INSERT INTO recommendation (user_id, song, artist, album, linkitunes, artwork, rating)";
$sql = $sql . "VALUES (".$_SESSION['id'].", '$song', '$artist', '$album', '$linkitunes', '$artwork', '$rating');";
        print "Hello.";
        $result = mysqli_query($conn,$sql) or die("Fail");
        ob_flush();
?>

Answer 3

您在PHP中没有任何转换为HTML的行...它将如何在屏幕上显示任何内容。在代码末尾写下这个。

echo "<br> Inserted Successfully";

mysqli_query的语法错误。使用以下一个。

$result=mysqli_query($conn,$sql) or die ("Fail");

其他可能的支票：

首先检查您的用户名和密码，并使用这段代码检查与数据库的连接是否成功。

if (!$conn) { 
    die('Could not connect to MySQL: ' . mysql_error()); 
}

使用session_start（）作为代码的第一行，否则您的会话将无效。这不是必需的，但可能会在将来导致问题。

mysqli_query（）上的空白页;？

3 个答案: