php提取MYSQL数据的url_php-通过URL中的特定ID从MySQL数据库获取数据

你的名字 2022-10-22 10:59 252阅读 0赞

我是PHP MySQL的新手.我正在开发一个歌本网站.

我正在尝试从URL site / publicsong.php？id = 12中的ID提取数据库中的数据.

// 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 “
“;

echo $row[‘eng_title’];

echo $row[‘eng_lyrics’];

echo $row[‘alphabet’];

}

?>

我将mysql_更改为mysqli_并添加了$conn.

而且我的结果还是空白.请帮助大家.提前致谢.

解决方法:

因此,我将通过一些修复来坚持您所拥有的.

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 “
“;

echo $row[‘id’];

echo $row[‘eng_title’];

echo $row[‘eng_lyrics’];

echo $row[‘alphabet’];

}

?>

标签：mysql,php

来源： https://codeday.me/bug/20191119/2037979.html

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