我是PHP MySQL的新手。我正在开发一个歌集网站。
我尝试从网址site/publicsong.php?id=12
中的ID提取数据库中的数据。
<?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。请帮帮我们提前谢谢。
答案 0 :(得分:5)
所以我会坚持你已有的一些修复。
<?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'];
}
?>
答案 1 :(得分:-1)
无需将id用作字符串,
您可以使用:id=" . $id;
答案 2 :(得分:-1)
试试这个:
<?php
$servername = "localhost";
$username = "";
$password = "";
$db_name = "test";
// Create connection
$conn = new mysqli($servername, $username, $password, $db_name);
// Check connection
mysql_select_db($db_name);
$id = $_GET['id'];
$id = mysql_real_escape_string($id);
$query = "SELECT * FROM `lyrics` WHERE `id`='" . $id . "'";
$result = mysql_query($query);
while($row = mysql_fetch_array( $result )) {
echo "<br><br>";
echo $row['title'];
echo $row['content'];
}
?>