Question

我有一个LOGIN PAGE，在我的数据库中有一个表，其成员包括ID，PASSWORD，USER_TYPE。我想要我的登录页面，当用户输入他们的ID和密码时，根据他们是谁来访问网站。（如果他们是学生去学生的页面）（如果他们是组织者，请访问组织者的页面）。

我无法通过以下编码获得结果：

<?php

        if ($_SERVER["REQUEST_METHOD"] == "POST")
{

        $user="admin";
        $pass="neehahs";
        $host="localhost";
        $db="login";

         $con=mysqli_connect($host,$user,$pass,$db);
                  if(mysqli_connect_errno($con)){
       echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }


        $username=($_POST['username']);
        $password=md5($_POST['password']);

        $username = mysqli_real_escape_string($con,$username);
        $password = mysqli_real_escape_string($con,$password);




        $sql="SELECT * FROM members WHERE student_id='%$username%' AND student_pass='%$password%'";
        $sqldata=mysqli_query($con,$sql)
        or die ("error");

        while ($row=mysqli_fetch_array($sqldata)){

        if($row["user_type"]=='student'){
    header('location: http://localhost/greenstudio/index.html');

        }

elseif
    ($row["user_type"]=='organizer'){
    header('location: http://localhost/greenstudio/index2.html');

    }else {
        echo"Sorry, your credentials are not valid, Please try again.";


    }
    }
    exit();     
        }

          ?>

Answer 1

你应该有一个类似于：

的表

Table: users
--------+----------+----------+----------
user_id | username | password | user_type
--------+----------+----------+----------
1       | admin    | neehahs  | organizer
2       | student1 | mypass   | student

然后您可以编写如下查询：

SELECT
  user_type
FROM
  users
WHERE
  BINARY username='$username' AND
  BINARY password='$password'

然后你的if：else if：else语句只会重定向返回是学生还是组织者;并且没有返回的行等于无效登录。

注意：使用BINARY与登录进行比较以确保用户名输入区分大小写，并且您应该在密码字段md5上使用某种加密，但强烈建议使用更强的加密

编辑：以下是我编写此逻辑的方法：

<?php if ($_SERVER["REQUEST_METHOD"] == "POST"){ $user="admin"; $pass="neehahs"; $host="localhost"; $db="login"; $con=mysqli_connect($host,$user,$pass,$db); if(mysqli_connect_errno($con)){ echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $username=($_POST['username']); $password=md5($_POST['password']); $username = mysqli_real_escape_string($con,$username); $password = mysqli_real_escape_string($con,$password); $sql="SELECT user_type FROM members WHERE BINARY student_id='$username' AND BINARY student_pass='$password'"; $sqldata=mysqli_query($con,$sql) or die ("error"); $row = mysqli_fetch_array($sqldata); if(is_null($row) || mysqli_num_rows($sqldata)!=1){ echo "Sorry, your credentials are not valid or matches more than 1 user, Please try again."; } else if(isset($row["user_type"])){ if($row["user_type"]=='student'){ header('location: http://localhost/greenstudio/index.html'); } else if($row["user_type"]=='organizer'){ header('location: http://localhost/greenstudio/index2.html'); } else { echo "User type was returned as not student nor organizer."; } } else { echo "Sorry, user_type was not returned in the dataset retrieved from the database."; } } ?>

将输入与两个表进行比较

1 个答案: