工作灯 - 如何检查，如果客户端已经登录，然后通过登录界面 [英] Worklight - How to check if a client is already logged in, then pass the login screen

问题描述

我收到以下错误，当我登录我的登录屏幕的第二次。

I get the following error when I log in the second time in my login screen.

    [ERROR   ] FWLSE0099E: An error occurred while invoking procedure  [project Klappr]KlapprAuthAdapter/submitAuthenticationFWLSE0100E:  parameters: [project Klappr]{
       "arr": [
  "dGVzdDp0ZXN0"
   ]
  }
   Cannot change identity of an already logged in user in realm 'KlapprAuthRealm'. The application must logout first.
    FWLSE0101E: Caused by:  [project Klappr]null 
                                                                                                           com.worklight.common.log.filters.ErrorFilter

在submitauthentication（在适配器），我检查，如果用户名和密码是否正确，如果正确设置我这样在WorklightRealm的activeUser：

In submitauthentication (in the adapter) I check if the username and password are correct and if they are correct I set the activeUser in the WorklightRealm like this:

    WL.Server.setActiveUser("KlapprAuthRealm",{
                        userId:""+teacher.id,
                        displayName: teacher.voornaam,
                        credentials:loginstring,
                        attributes: {
                            "teacherId": teacher.id,
                        }
                    });

我如何检查，如果用户已经登录？这是最好的客户端之前，我展示loginpage？或者我应该让他们能够登录在另一个时间，如果他们已经登录，也返回用户id？

How can I check if the user is already logged in? Is this best on client side before I show the loginpage? Or should I let them be able to log in another time, and if they are already logged in, also return the userId?

推荐答案

这是发生，因为你想设置为有效用户，但它已经被设置。一个可能的解决办法是设置活动的用户以实际用户身份之前调用WL.Server.setActiveUser（KlapprAuthRealm，NULL）。

This is happening because you're trying to set active user but it is already set. A possible solution might be to call WL.Server.setActiveUser("KlapprAuthRealm", null) before setting active user with actual user identity.

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