您好我想知道为什么code不起作用。我想查询ParseUser的用户名现场找到一个特定的用户，但它口口声声说，它不能找到它。

Hi I was wondering why this code does not work. I am trying to query a ParseUser's username field to find a certain user but it keeps saying that it cant find it.

private void findUserName(String user) {
        // query the User database to find the passed in user
        ParseQuery query = ParseUser.getQuery();
        query.whereEqualTo("username", user);
        query.findInBackground(new FindCallback() {
            @Override
            public void done(List<ParseObject> objects, ParseException e) {
                foundUser = (objects.size() != 0);
            }
        });
    }

下面是我的方法调用它

if (!foundUser) {
    errorMessage.setText("Invalid user name");
}

foundUser是一个字段，因为我不可能在该方法返回它...

foundUser is a field because I couldnt return it in the method...

推荐答案

解析会将用户从解析对象的对象区分开来。您应该使用名单，其中，ParseUser＆GT; ，而不是的名单，其中，的parseObject＆GT; 。解析Android的指南提供了一个例子https://parse.com/docs/android_guide#users-querying. 下面是解析例如你的where子句。

Parse treats User objects separate from Parse objects. You should use List<ParseUser> instead of List<ParseObject>. The Parse Android Guide provides an example https://parse.com/docs/android_guide#users-querying. Here is the Parse example with your where clause.

ParseQuery<ParseUser> query = ParseUser.getQuery();
query.whereEqualTo("username", user);
query.findInBackground(new FindCallback<ParseUser>() {
  public void done(List<ParseUser> objects, ParseException e) {
    if (e == null) {
        // The query was successful.
    } else {
        // Something went wrong.
    }
  }
});