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从异步任务返回ArrayList

[英]Return ArrayList from async task

1) I already read all questions about that here, so please tell me more, than just url from SO... 1)我已经在这里阅读了有关此问题的所有问题,因此,请告诉我更多信息,而不仅仅是SO的网址...

2) I want to return only ONE variable from asynctask, but it can be message error or user details. 2)我只想从asynctask返回一个变量,但是它可能是消息错误或用户详细信息。 So I decided to return ArrayList with indexes 0 (for error messages) and 1 (for success). 因此,我决定返回索引为0(错误消息)和1(成功)的ArrayList。

AsyncTask 异步任务

public class LoginGet extends AsyncTask<String, ProgressBar, ArrayList<String>> {
...
public ArrayList<String> result = new ArrayList<String>();

public LoginGet(Context mContext, String loginText, String passwordText) {
    //Get values form LoginActivity
}

protected void onPreExecute(){
    super.onPreExecute();
    // progressDialog
}

@Override
protected ArrayList doInBackground(String... arg0){
    try {
        // Making HTTP Request
        try {
            // Creating HTTP client
            ...
            ...
            JSONObject jObject = new JSONObject(json);
            success = jObject.getString("success");
            result.add(0, "");
            result.add(1, "");
            if (success != null) {
                if (success == "1") {
                    result.add(1, jObject.getString("hash"));
                    return result;
                } else if (success == "0") {
                    result.add(0, jObject.getString("error"));
                    return result;
                }
            } else {
                result.add(0, "Error while logging.");
                return result;
            }

           // writing response to log
            Log.d("Http Response:", response.toString());
        }  catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }


    }catch (Exception e) {

    }

    return result;
}

@Override
protected void onPostExecute(ArrayList<String> result) {
    ((LoginActivity)context).getResult(result);
    progressDialog.dismiss();
}
}

LoginActivity 登录活动

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    ...

    loginButton.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            ...
            if (loginText.compareTo("") == 0 || passwordText.compareTo("") == 0) {
                ...
            }else{
                LoginGet loginPost = new LoginGet(LoginActivity.this,loginText,passwordText);
                loginPost.execute();
            }
        }
    });

}

// Get result from asynctask LoginGet
public void getResult(ArrayList<String> result){
    if (result.get(0) != "") {
        loginButton.setEnabled(true);
        Toast.makeText(LoginActivity.this, result.get(0), Toast.LENGTH_SHORT).show();
    } else if (result.get(1) != "") {
        SharedPreferences preferences = PreferenceManager.getDefaultSharedPreferences(LoginActivity.this);
        SharedPreferences.Editor editor = preferences.edit();
        editor.putString("Login", result.get(1));
        editor.commit();

        Intent intent = new Intent(LoginActivity.this, MainActivity.class);
        startActivity(intent);
    }

}

Error: 错误:

java.lang.IndexOutOfBoundsException: Invalid index 0, size is 0
at java.util.ArrayList.throwIndexOutOfBoundsException(ArrayList.java:255)
at java.util.ArrayList.get(ArrayList.java:308)
at cz.jacon.smsapp.LoginActivity.getResult(LoginActivity.java:57)
at cz.jacon.smsapp.background.LoginGet.onPostExecute(LoginGet.java:121)
at cz.jacon.smsapp.background.LoginGet.onPostExecute(LoginGet.java:30)

3) In my AsyncTask I have result.add(0, ""); 3)在我的AsyncTask中,我有result.add(0, ""); and result.add(1, ""); result.add(1, ""); , so why is the ArrayList size 0? ,为什么ArrayList的大小为0?

In my opinion result.add(0, ""); 我认为result.add(0, ""); isn't even reached, since the try block might be failing. 甚至没有达到,因为try块可能失败了。

And sadly you have an empty catch block, just waiting to skip them errors: 不幸的是,您有一个空的catch块,只是在等待跳过它们的错误:

catch (Exception e) {

}

Your solution would be to add these 您的解决方案是添加这些

result.add(0, "");
result.add(1, "");

to the very beginning of the doInBackground or even to the constructor and you should be fine. doInBackground的最开始甚至是构造函数,您都应该可以。

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