Question

我创建了一个类并使用AsyncTask进行扩展，现在我想将一个字符串返回给另一个活动抛出这个类。它从mysql mDatabase获取jsonstring

enter code here  

public class GetJson  extends AsyncTask<Void,Void,String> {
String  JSON_STRING;
String json_url;
public String pasString;
Activity ab;
@Override
protected void onPreExecute() {
    json_url="https://XXXXXXX.000webhostapp.com/Json_getData_Survey.php";
}

public GetJson(Activity b) {
    ab=b;
}

@Override
protected String doInBackground(Void... params) {

    try {
        URL url =new URL(json_url);
        HttpURLConnection httpURLConnection =(HttpURLConnection)  url.openConnection();
        InputStream inputStream  =httpURLConnection.getInputStream();
        BufferedReader bufferedReader=new BufferedReader(new InputStreamReader(inputStream));
        StringBuilder stringBuilder=new StringBuilder();

        while ((JSON_STRING=bufferedReader.readLine())!=null)
        {

            stringBuilder.append(JSON_STRING+"\n");
        }


        bufferedReader.close();
        inputStream.close();
        httpURLConnection.disconnect();

        String  strings =stringBuilder.toString().trim();
        return  strings;

    } catch (MalformedURLException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    return null;
}

@Override
protected void onProgressUpdate(Void... values) {
    super.onProgressUpdate(values);
}

@Override
protected void onPostExecute(String result) {

    pasString=result;

   // Toast.makeText(ab,pasString,Toast.LENGTH_LONG).show();
   // json_string1=result;
}

}

我只需要将string1返回到我的另一个Activity，我已经在GetJson类中创建了另一个mehtod来返回此字符串但是当我显示它时它没有显示任何内容。请帮忙

Answer 1

非常简单创建一个接口类如下

public interface ResponseListener() {
      void onResponseReceive(String data);
}

并在你的GetJson类中添加此类的实例

public class GetJson  extends AsyncTask<Void,Void,String> {
String  JSON_STRING;
String json_url;
public String pasString;
Activity ab;

// this is new code
ResponseListener listener;
public void setOnResponseListener(ResponseListener listener) {
    this.listener = listener;
}

@Override
protected void onPreExecute() {
    json_url="https://XXXXXXX.000webhostapp.com/Json_getData_Survey.php";
}

// now add this code in onPost function

@Override
protected void onPostExecute(String result) {

   pasString=result;
   listener.onResponseReceive(pasString);

  // Toast.makeText(ab,pasString,Toast.LENGTH_LONG).show();
  // json_string1=result;
}

现在当你从你的活动中调用GetJson时，只需简单地做到这一点;

GetJson json = new GetJson(Activity.this);
json.setOnResponseListener(new ResponseListener() {
       @Override
       public void onResponseReceived(String data) {
           // here you will get your response                                               
       }
});

希望你能理解。：）

Answer 2

~~尝试使用活动结果。您可以在一个活动上设置返回值，并在该活动终止时获取它。~~

~~https://developer.android.com/training/basics/intents/result.html~~

如何将字符串从AsyncTask返回到另一个活动

2 个答案: