doulang6695 2012-01-04 08:31 采纳率: 100%
浏览 65
已采纳

如何在iPhone应用程序中执行注销?

I know nothing about PHP, but in my application I need to implement user login and logout. I've found this nice tutorial and have successfully implemented logging in, but how do I log out? I just can't find any information about it in google, particularly about iPhone side. PHP tutorials mostly explain a lot of theory about sessions and stuff like that which is not applicable for my specific event. I would love to spend my time for learning that theory, but I have to learn PHP basics first and this is not what I currently need.

All I need is to understand the correct way of implementing logout.

  1. I would be glad very much if someone explained me shortly in plain language the mechanism of standard interaction between iPhone app and php script when logging in and out.
  2. This is more important. Would you please provide me with few lines of php and objective- с code displaying how to implement logging out?
  • 写回答

1条回答 默认 最新

  • douzhong2954 2012-01-04 09:09
    关注

    In our app we use this schema:

    Login I'll jumpover (that you've done)

    When the user logout our application call an URL, which is considered for logout. For example: http://ourcoresyste.com/logout.php

    Sorry for inconvinience, but I'm not iOS developer, so I don't know exactly how application makes call, but I think you should know it, if login is done.

    In the script logout.php:

    <?php
        if (!isset($_SESSION))
        {
            exit(json_encode(array('code' => 0, 'message' => 'Logout successful')));
        }
        esle
        {
            $user = $_SESSION['user_id']; // suppose you have stored into $_SESSION['user_id'] logged in user;
            // Do some stuff while logout, maybe some DB interactions.
            if ($shit_happens)
            {
                exit(json_encode(array('code' => 1, 'message' => 'Shit happens')));
            }
            unset($_SESSION['user_id']); // or session_destroy(), if you want completly remove all information about user.
            exit(json_encode(array('code' => 0, 'message' => 'Logout successful')));
        }
    

    Than in your app you parse JSON response which will be like this:

    {"code":0,"message":"Your detailed message"}
    

    And decide by code retrieved from JSON response, were user logged out or not. If he was, next time you call any PHP script, which relies on $_SESSION['user_id'], will fail, probably that's mean that user is logged out.

    I think that's all.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

    报告相同问题?

    悬赏问题

    • ¥15 Qt 不小心删除了自带的类,该怎么办
    • ¥15 我需要在PC端 开两个抖店工作台客户端.(语言-java)
    • ¥15 有没有哪位厉害的人可以用C#可视化呀
    • ¥15 可以帮我看看代码哪里错了吗
    • ¥15 设计一个成绩管理系统
    • ¥15 PCL注册的选点等函数如何取消注册
    • ¥15 问一下各位,为什么我用蓝牙直接发送模拟输入的数据,接收端显示乱码呢,米思齐软件上usb串口显示正常的字符串呢?
    • ¥15 Python爬虫程序
    • ¥15 crypto 这种的应该怎么找flag?
    • ¥15 代码已写好,求帮我指出错误,有偿!