PHP登录脚本只返回一个条件

Question

/ *当我测试登录页面时，我输入我在phpMyAdmin中手动设置的用户帐户的信息，但它一直返回错误，＆＃34;找不到用户名。＆＃34; * /

<?php
 session_start();
  include_once '../../config/dbconfig.php';

if(isset($_SESSION['user'])!="")
{
    header("Location:http://www.google.com");
}
if(isset($_POST['btn-login']))
{
  $email = mysqli_real_escape_string($con,$_POST['email']);
  $upass = mysqli_real_escape_string($con,$_POST['pass']);
  $query= "SELECT email, pass from  users WHERE email='$email' and pass  =$upass'";
  $result= mysqli_query($con,$query);

if($result)
   {
      echo "user found";
   }
 else
{
    echo "user was not found";
  }
}

Answer 1

缺少引语：

$query= "SELECT email, pass from  users WHERE email='$email' and pass  =$upass'";

应该是：

$query= "SELECT email, pass from  users WHERE email='$email' and pass  = '$upass'";

引用是在$upass

之前添加的