java - 解析 Android 中出现问题的查询

Question

嘿，大家好，我正在尝试使用解析查询从解析数据库中检索一个值，但是当我尝试从检查表中特定值是否存在的方法返回 boolean 值时，我不太理解程序的流程。这是代码:

public class UsersSignUp extends Activity {

Button btnxxx;
EditText edt;
Context mContext = this;
CheckInternetConnectivity checkInternetConnectivity;
String housenumber;
String objectid;
boolean b;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_xyyxyx);

    btnxxx = (Button) findViewById(R.id.btnSignUp);
    btnxxx.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            edt = (EditText) findViewById(R.id.txtHouse);
            if (isValidHouseNum(edt.getText().toString())) {
                Log.d("Bool Value 2:", edt.getText().toString());
                String h=edt.getText().toString();
                Log.d("Bool Value 2.1:", h);
                b=checkIfRegistered(h);
                Log.d("Check: ", String.valueOf(b));
            }


            else {
                Animation shake = AnimationUtils.loadAnimation(UsersSignUp.this, R.anim.shake);
                edt.startAnimation(shake);
                Toast.makeText(getApplicationContext(), "Please enter valid house number!", Toast.LENGTH_LONG).show();
            }
        }
    });
}

public boolean isValidHouseNum(String housenumber) {
    boolean check;
    if (housenumber.length() > 6 || housenumber.length() < 6) {
        check = false;
    } else {
        check = true;
    }

    return check;
}


public boolean checkIfRegistered(String house_number)
{
    ParseQuery<ParseObject> query2=ParseQuery.getQuery("SampleTable_123456");//Table Name in parse
    query2.whereEqualTo("housenumber", house_number);
    query2.getFirstInBackground(new GetCallback<ParseObject>() {
        @Override
        public void done(ParseObject parseObject, ParseException e) {
            if (parseObject == null) {
                Log.d("Message 1:", "Number not retreived");
                b = false;
                Log.d("Message 4:", String.valueOf(b));
            } else {
                Log.d("Message 2:", "Number Retrieved");
                String number = parseObject.getString("housenumber");
                b = true;
                Log.d("Message 5:", String.valueOf(b));
                Log.d("Message 2:", number);
            }
        }
    });

    Log.d("Message 3:", String.valueOf(b));
    return b;
}

这是当我搜索的号码存在于数据库中时我收到的日志:

• 值2::123456
• 值2.1:: 123456
• D/消息 3:: false
• D/检查:: false
• D/消息2::已检索号码
• D/消息 5:: true
• D/消息2:: 123456

虽然它应该首先返回 true，但在看到日志后它已经存在，但仍然非常困惑。帮帮我吧伙计们。

Answer 1

如果您调试此代码，您会发现在调用 getFirstInBackground 后，它将跳转到 Log.d("Message 3:", String.valueOf(b)) 语句，此时其 false

不返回任何值将您想要的任何内容放入done

query2.getFirstInBackground(new GetCallback<ParseObject>() {
    @Override
    public void done(ParseObject parseObject, ParseException e) {
        if (parseObject == null) {
           //null means number is not registered
          // next step must be here if number not register dont return any value



        } else {
           //number is registerd so print toast here
        }
    }
});