解析：无效的用户名，密码 [英] Parse : invalid username, password

问题描述

我是使用解析API用于数据库，并试图使用的用户名服务，它提供。据我所知，从教程，为了登录你这样做：

I was using the Parse API for databases and trying to use the username service that it provides. I understand that from the tutorial that in order to login you do this :

ParseUser.logInInBackground("Jerry", "showmethemoney", new LogInCallback() {
  public void done(ParseUser user, ParseException e) {
    if (user != null) {
      // Hooray! The user is logged in.
    } else {
      // Signup failed. Look at the ParseException to see what happened.
    }
  }
});

如果登录失败，我只是想知道我怎么能知道它是否失败，因为在键入的用户名无效或密码。我知道，你可以做e.get code（），以获得错误的发生的类型，但是从本网站的 https://parse.com/docs/android/api/ 我找不到任何错误codeS属于无效的用户名/密码

If the login failed, I was just wondering how I could tell whether it failed because the username typed in was invalid, or the password. I know that you can do e.getCode() to get the type of error that occurred, but from this site https://parse.com/docs/android/api/ I couldn't find any error codes pertaining to invalid username/password

感谢您
詹姆斯

Thank you james

推荐答案

这是我做过什么检查用户名/密码有效期在我的应用程序之一。此法提交反对ParseUser类的查询，如果传递的用户名存在返回true，如果它不那么你知道用户名是有效的。

This is what I did to check username/password validity in one of my applications. This method submits a query against the ParseUser class and returns true if the passed username exists, if it does then you know the username is valid.

（外部检查ParseException.OBJECT_NOT_FOUND - 结合这一点，我们可以告诉用户是否需要注册或有一个无效的密码）

(check externally for ParseException.OBJECT_NOT_FOUND - in conjunction with this we can tell whether the user needs to register or has an invalid password.)

public boolean queryCredentials(String username) {
        ParseQuery<ParseUser> queryuserlist = ParseUser.getQuery();
        queryuserlist.whereEqualTo("username", username);
        try {
            //attempt to find a user with the specified credentials.
            return (queryuserlist.count() != 0) ? true : false;
        } catch (ParseException e) {
            return false;
        }
    }

希望这可以帮助别人这个问题。

Hopefully this can help someone with this issue.

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