从内部访问私有变量 class (AsyncTask)

Access private variable from inner class (AsyncTask)

我有问题,我正在尝试从 AsyncTaskdoInBackground 方法访问外部 class 的私有变量,该方法是内部 class .这里有完整的代码。

public class MessageHandler {
private Contact receiver;
private String senderFacebookId;
private String message;
private final String TAG = "MessageHandler";
private Context context;
boolean sentResult;

public MessageHandler(Contact receiver, String senderFacebookId, String message, Context context) {
    this.receiver = receiver;
    this.senderFacebookId = senderFacebookId;
    this.message = message;
    this.context=context;
    this.sentResult = false;
}

public void send(){
    Log.i(TAG, "Sending message to " + receiver.getName() + " receiver id: " + receiver.getFacebook_id() + " sender id: " + senderFacebookId + " message: " + message);

    new SenderAsync().execute(senderFacebookId,message, receiver.getFacebook_id());
    if(this.sentResult==true){
        Toast toast = Toast.makeText(this.context, "Message Sent", Toast.LENGTH_LONG);
        toast.show();

    }else{
        Toast toast = Toast.makeText(this.context, "ERROR: Message not Sent", Toast.LENGTH_LONG);
        toast.show();
    }
}



private class SenderAsync extends AsyncTask<String,String,String> {

    @Override
    protected void onPreExecute(){
        super.onPreExecute();
    }

    @Override
    protected void onPostExecute(String s) {
        super.onPostExecute(s);
    }

    @Override
    protected String doInBackground(String... params) {
        HttpClient client = new DefaultHttpClient();

        try {
            List<NameValuePair> parameter = new ArrayList<>(1);
            parameter.add(new BasicNameValuePair("regId", "empty"));
            parameter.add(new BasicNameValuePair("sender_id", params[0]));
            parameter.add(new BasicNameValuePair("receiver_facebook_id", params[2]));

            parameter.add(new BasicNameValuePair("message", params[1]));

            String paramString = URLEncodedUtils.format(parameter, "utf-8");
            HttpGet get = new HttpGet(ProfileFragment.SERVER_URL_SEND_MESSAGE+"?"+paramString);

            Log.i(TAG, paramString);
            HttpResponse resp = client.execute(get);
            System.out.println("SREVER RESPONSE " + resp.getStatusLine().getStatusCode());

            //THIS IS THE VARIABLE THAT I WANT TO ACCESS
            if(resp.getStatusLine().getStatusCode()==200){
                MessageHandler.this.sentResult = true;
            }



        } catch (IOException e) {

            Log.i(TAG,"Error :" + e.getMessage());
        }
        return null;
    }
}

}

我试图访问的变量是布尔变量sentResult。我还打印出始终为 200 的服务器的响应。即使我删除了 if 条件并将它设置为 true 在任何情况下,看起来那行代码从未执行并且未访问变量,所以它总是错误的。

为 SenderAsync 创建构造函数;

public SenderAsync(上下文上下文,字符串发送者 ID,字符串消息,联系接收者)

这样称呼它;

new SenderAsync(senderFacebookId,message, receiver.getFacebook_id()).execute();

并且您需要在 onPostExecute 中获取结果。 void onPostExecute(字符串返回值) { // 运行 这里是你想要的 if 语句 }

异步任务的全部意义在于它是异步的。如果您执行一个任务,您将不会在下一行代码中得到结果。为了评估结果,您应该使用 onPostExecute 回调方法。

@Override
protected void onPostExecute(String s) {
    if(sentResult==true){
        Toast toast = Toast.makeText(this.context, "Message Sent", Toast.LENGTH_LONG);
        toast.show();
    }else{
        Toast toast = Toast.makeText(this.context, "ERROR: Message not Sent", Toast.LENGTH_LONG);
        toast.show();
    }
}

此外,如果您要更改方法签名,您甚至不需要外部变量。

private class SenderAsync extends AsyncTask<String,String,Boolean> {

@Override
protected void onPostExecute(Boolean sent) {
    if(sent == true){
        Toast toast = Toast.makeText(this.context, "Message Sent", Toast.LENGTH_LONG);
        toast.show();
    }else{
        Toast toast = Toast.makeText(this.context, "ERROR: Message not Sent", Toast.LENGTH_LONG);
        toast.show();
    }
}

@Override
protected Boolean doInBackground(String... params) {
    try {
        ...
        if (resp.getStatusLine().getStatusCode() == 200) {
            return true;
        }
    }
    catch (IOException e) {

    }

    return false;
}

您正在尝试在告诉 AsyncTask 完成其工作后立即验证 sentResult。 AsyncTask 在单独的线程上并行执行它必须执行的操作。所以在它完成之前,sentResult 是不变的。

我的建议是将您的 Toast 逻辑放在 onPostExecute 中。

您没有正确使用 AsyncTask。 AsyncTask 应该 return 结果 onPostExecute() 并且你的 UI 应该有一个单独的方法 onPostExecute() 调用(因此异步名称)。对于你想要完成的事情,你应该这样做:

public void send() {
    new SenderAsync().execute(senderFacebookId,message, receiver.getFacebook_id());
}

public void processResult(boolean result) {
  String resultMsg = result ? "Message Sent" : "ERROR: Message not Sent";
  Toast.makeText(this.context, resultMsg, Toast.LENGTH_LONG).show();
}

private class SenderAsync extends AsyncTask<String, String, Boolean> {
  ...
  @Override
    protected String doInBackground(String... params) {
      ...
      try {
        ...
        return (resp.getStatusLine().getStatusCode() == 200)
      }
      ...
      return false;
    }

    @Override
    protected void onPostExecute(Boolean result) {
        processResult(result);
    }
}