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Parse errors are not displayed
(8个答案)
4年前关闭。
我正在建立一个基本的歌曲推荐站点,并且建立了一个表单,该表单会导致页面上包含以下确切代码:
它始终显示“你好”。字符串,直到添加$ sql值。我认为代码语法有问题,但不确定。尝试了很多变化。以防万一,我还添加了表单代码:
(8个答案)
4年前关闭。
我正在建立一个基本的歌曲推荐站点,并且建立了一个表单,该表单会导致页面上包含以下确切代码:
<?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();
?>
它始终显示“你好”。字符串,直到添加$ sql值。我认为代码语法有问题,但不确定。尝试了很多变化。以防万一,我还添加了表单代码:
<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>
最佳答案
你应该执行这个...
$result = mysqli_query($con,$sql); or die("Fail");
关于php - mysqli_query()上的空白页? ,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/28493096/
10-13 00:59