Question

我希望在没有Object的情况下获得Parse中something.put("blalala")...的字符串。

仅从Parse获取字符串。我的代码出了什么问题？我怎么能这样做？

public class MyActivity extends Activity {


TextView textView;

 @Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_my);
    Parse.initialize(this, "blablablbalba", "blablablabla");

    final ParseObject gameScore = new ParseObject("GameScore");

     //txtv
     textView = (TextView)findViewById(R.id.textView);
     //txtv


     ParseQuery<ParseObject> query = ParseQuery.getQuery("GameScore");

     query.getInBackground("theid", new GetCallback<ParseObject>() {
         @Override
         public void done(ParseObject parseObject, com.parse.ParseException e) {
             if (e == null){

                 String playerName = gameScore.getString("playerName");
                 textView.setText(playerName);

             }
             else {
                 textView.setText("Mince !!!");
             }
         }
     });







 }

}

Answer 1

代码应该是String playerName = parseObject.getString("playerName"); parseObject是保存查询结果的变量，而不是gameScore变量

（编辑:)直接来自文档：

    ParseQuery<ParseObject> query = ParseQuery.getQuery("GameScore");
query.getInBackground("xWMyZ4YEGZ", new GetCallback<ParseObject>() {
  public void done(ParseObject object, ParseException e) {
    if (e == null) {
      // object will be your game score
    } else {
      // something went wrong
    }
  }
});

从Parse中检索数据

1 个答案: