php - 从 MySQL 更新到 MySQLi 时出现从数据库中提取数据的问题

Question

我希望将其从 MySQL 更新到 MySQLi，但我在这样做时一直遇到错误。基本上，这是从另一个页面上的链接创建的动态页面的一段代码。

$connect = mysql_connect('localhost', 'root', 'Password');
$select_db = mysql_select_db('playerslog');

$id = mysql_real_escape_string($_GET['UUID']);
//Remove LIMIT 1 to show/do this to all results.
$query = 'SELECT * FROM `playerslog` WHERE `UUID` = '.$id.' LIMIT 1';
$result = mysql_query($query);
$row = mysql_fetch_array($result);
// Echo page content

关于如何完成此任务有什么建议吗？感谢您的宝贵时间!

按要求更新。

<?php
$connect = mysqli_connect('localhost', 'root', 'Password');
$select_db = mysqli_select_db('playerslog');


$id = mysqli_real_escape_string($_GET['UUID']);
//Remove LIMIT 1 to show/do this to all results.
$query = 'SELECT * FROM `playerslog` WHERE `UUID` = '.$id.' LIMIT 1';
$result = mysqli_query($query);
$row = mysqli_fetch_array($result);
// Echo page content

?>

Answer 1

您不能简单地将 mysql_* 替换为 mysqli_*。它们有不同的语法。例如，您应该修复执行查询的方式。 Mysqli 需要两个参数:连接和查询。您只传递查询:

<?php
$connect = mysqli_connect('localhost', 'root', 'Password');
$select_db = mysqli_select_db('stats');
$id = mysqli_real_escape_string($_GET['UUID']);
//Remove LIMIT 1 to show/do this to all results.
$query = 'SELECT * FROM `playerslog` WHERE `UUID` = '.$id.' LIMIT 1';
$result = mysqli_query($connect, $query);
$row = mysqli_fetch_array($result);
// Echo page content

?>