我正在尝试创建一个登录页面。 连接成功但查询未获取所需结果。 登录页面通过post方法向action_page.php发送数据。
Action_page.php
<?php
$servername = "localhost:3306";
$username = "root";
$password = "";
$dbname = "MyDB";
$con = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$uname = $_POST['uname'];
$pass = $_POST['psw'];
$sql = "SELECT * FROM Users WHERE uname=".$uname." AND pass=".$pass;
echo "<br/>$sql<br/>";
//$result = $con->query($sql);
echo var_dump($con->query($sql));
$con->close();
有人可以帮忙吗!
最佳答案
You can use single quote for the inserted variable inside double quote :)
$sql = "SELECT * FROM Users WHERE uname='$uname' AND pass='$pass'";
关于php - 已建立 MySQL 连接但查询失败,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47688616/