我是 PHP MySQL 新手。我正在开发一个歌曲集网站。
我正在尝试从 URL 中的 ID 提取数据库中的数据site/publicsong.php?id=12
.
<?php
// Create connection
$conn = new mysqli($servername = "localhost";$username = "dansdlpe_dan";$password = "g+GbMr}DU4E@";$db_name = "dansdlpe_lyrics";);
// Check connection
$db_name = "dansdlpe_lyrics";
mysqli_select_db($conn,$db_name);
$id = $_GET['id'];
$id = mysqli__real_escape_string($conn,$id);
$query = "SELECT * FROM `lyrics_a` WHERE `id`='" . $id . "'";
$result = mysqli__query($conn,$query);
echo $row['id']; while($row = mysqli__fetch_array( $result )) {
echo "<br><br>";
echo $row['eng_title'];
echo $row['eng_lyrics'];
echo $row['alphabet'];
}
?>
我将 mysql 更改为 mysqli 并添加了 $conn。
我的结果仍然是空白。请大家帮忙。提前致谢。
因此,我将坚持您已有的内容并进行一些修复。
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
$servername = "localhost";
$username = "xxxx";
$password = "xxxxxx";
$db_name = "xxxxxxxxx";
// Create connection
$conn = new mysqli($servername, $username, $password, $db_name);
// Check connection
if ($conn->connect_error){
die("Connection failed: " . $conn->connect_error);
}
$id = $_GET['id'];
$id = mysqli_real_escape_string($conn,$id);
$query = "SELECT * FROM `lyrics_a` WHERE `id`='" . $id . "'";
$result = mysqli_query($conn,$query);
while($row = mysqli_fetch_array($result)) {
echo "<br><br>";
echo $row['id'];
echo $row['eng_title'];
echo $row['eng_lyrics'];
echo $row['alphabet'];
}
?>
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