Question

一直在尝试从 Parse.com 中检索特定用户，并在片段中使用对象ID 。从调试器看来，findinBackground，getInBackground或getFirstinbackground的代码似乎永远不会被执行。

我的 getFirstInBackground 方法（完整代码）

package com.example.jamesytl.hostel;

import android.app.Activity;
import android.content.Intent;
import android.graphics.Bitmap;
import android.graphics.BitmapFactory;
import android.os.Bundle;
import android.support.v4.app.Fragment;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import android.widget.ImageView;
import android.widget.TextView;
import android.widget.Toast;

import com.parse.FindCallback;
import com.parse.GetCallback;
import com.parse.ParseException;
import com.parse.ParseFile;
import com.parse.ParseObject;
import com.parse.ParseQuery;
import com.parse.ParseUser;

import java.util.List;

/**
 * Created by jamesytl on 15/4/2016.
 */
public class hosteldetails extends Fragment{
    String ownerID,email,hostelno,roomtype,desc,status,area,name;
    int price;
    ImageView image;
    TextView txtname,txtemail,txthostelno,txtprice,txtroomtype,txtdesc,txtstatus,txtarea;
    hostelwithoutparse currenthostel;
    Bundle bundle;
    ParseUser owner;
    byte[] imgByte;
    public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
        View v = inflater.inflate(R.layout.hosteldetails, container, false);
        bundle = this.getArguments();
        if (bundle != null)
            currenthostel = bundle.getParcelable("item_selected_key");
        else
            Toast.makeText(getActivity().getBaseContext(), "Failed", Toast.LENGTH_SHORT).show();

        hostelno = currenthostel.getHostelNo();
        price = currenthostel.getHostelPrice();
        roomtype = currenthostel.getHostelRoom();
        desc = currenthostel.getHostelDescription();
        status = currenthostel.getHostelStatus();
        area = currenthostel.getHostelArea();
        ownerID = currenthostel.getOwner();
        imgByte = currenthostel.getImgByte();

        ParseQuery<ParseUser> query = ParseUser.getQuery();
        query.whereEqualTo("objectId", ownerID);
        query.getFirstInBackground(new GetCallback<ParseUser>() {
            @Override
            public void done(ParseUser object, ParseException e) {
                name = object.getUsername();
                email = object.getEmail();
            }
        });
        txtname = (TextView) v.findViewById(R.id.tvDOwnerName);
        txtarea = (TextView) v.findViewById(R.id.tvDArea);
        txtprice = (TextView) v.findViewById(R.id.tvDprice);
        txthostelno = (TextView) v.findViewById(R.id.tvDHostelNo);
        txtroomtype = (TextView) v.findViewById(R.id.tvDRoom);
        txtdesc = (TextView) v.findViewById(R.id.tvDDesc);
        txtstatus = (TextView) v.findViewById(R.id.tvDStatus);
        txtemail = (TextView) v.findViewById(R.id.tvDEmail);
        image = (ImageView) v.findViewById(R.id.imgDHostel);


        Bitmap bitmap = BitmapFactory.decodeByteArray(imgByte, 0, imgByte.length);
        image.setImageBitmap(bitmap);

        txtname.setText(name);
        txtemail.setText(email);
        txtarea.setText(area);
        txtprice.setText("RM" + String.valueOf(price));
        txthostelno.setText(hostelno);
        txtroomtype.setText(roomtype);
        txtdesc.setText(desc);
        txtstatus.setText(status);
        return v;
    }

}

有什么线索我怎样才能做到这一点？它与我在片段中运行它导致它无法在后台运行有关吗？

Answer 1

好的，到目前为止您的实际代码没有任何问题。根据您的评论，query.getFirstInBackground(new GetCallback<ParseUser>()部分确实已执行，您想知道为什么第二部分：public void done(ParseUser object, ParseException e)未执行。

原因是query.getFirstInBackground()是一个异步任务，这意味着它将在后台运行（在后台获取数据），当它完成提取时，它将执行public void done()中的代码。

因此，我的建议是您将代码 IN public void done()方法，以便执行 AFTER 完成提取。如下：

query.getFirstInBackground(new GetCallback<ParseUser>() {
        @Override
        public void done(ParseUser object, ParseException e) {
            name = object.getUsername();
            email = object.getEmail();

            Bitmap bitmap = BitmapFactory.decodeByteArray(imgByte, 0, imgByte.length);
            image.setImageBitmap(bitmap);

            txtname.setText(name);
            txtemail.setText(email);
            txtarea.setText(area);
            txtprice.setText("RM" + String.valueOf(price));
            txthostelno.setText(hostelno);
            txtroomtype.setText(roomtype);
            txtdesc.setText(desc);
            txtstatus.setText(status);
            return v;
        }
    });

无法在片段中使用objectID检索ParseUser

1 个答案: