java - Android 在 AsyncTask onPostExecute 中使用 Interface 返回值

Question

这个问题已经有答案了:

Android AsyncTask don't return correct Result (3 个回答)

已关闭 8 年前。

在下面的代码中，我想从 AsyncTask 返回值使用接口(interface)。但我得到了错误的值，并且无法从 onPostExecute 返回正确的值.

我开发了这个link教程与我的代码。我无法正确使用它。

界面:

public interface AsyncResponse {
    void processFinish(String output);
}

Ksoap主类:

public class WSDLHelper  implements AsyncResponse{
    public SoapObject request;

    private String Mainresult;

    public String call(SoapObject rq){

        ProcessTask p =new ProcessTask(rq);
        String tmp = String.valueOf(p.execute());

        p.delegate = this;

        return Mainresult;
    }


    @Override
    public void processFinish(String output) {

        this.Mainresult = output;
    }
}
class ProcessTask extends AsyncTask<Void, Void, Void > {
    public AsyncResponse delegate=null;

    SoapObject req1;
    private String result;
    public ProcessTask(SoapObject rq) {
        req1 = rq;
    }

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

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

        SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
        envelope.setOutputSoapObject(this.req1);

        AndroidHttpTransport transport = new AndroidHttpTransport(Strings.URL_TSMS);
        transport.debug = true;

        try {
            transport.call(Strings.URL_TSMS + this.req1.getName(), envelope);
            this.result = envelope.getResponse().toString();
        } catch (IOException ex) {
            Log.e("" , ex.getMessage());
        } catch (XmlPullParserException ex) {
            Log.e("" , ex.getMessage());
        }

        if (result.equals(String.valueOf(Integers.CODE_USER_PASS_FALSE))) {
            try {
                throw new TException(PublicErrorList.USERNAME_PASSWORD_ERROR);
            } catch (TException e) {
                e.printStackTrace();
            }

        }

        Log.e("------------++++++++++++++++-----------", this.result);

        return null;
    }

    protected void onPostExecute(String result) {
    /* super.onPostExecute(result);*/
        delegate.processFinish(this.result);
    }

}

请帮我解决这个问题

Answer 1

那是行不通的。您正在创建并执行 AsyncTask(异步!)，然后在尚未收到结果时调用 return Mainresult(同步!)。解决办法是去掉多余的类WSDLHelper，直接访问ProcessTask

除此之外，您错误地使用了 AsyncTask(将结果保存在类变量中，而不是将其作为参数传递)。这是完整版本:

public class ProcessTask extends AsyncTask<Void, Void, String> {
public AsyncResponse delegate=null;

SoapObject req1;

public ProcessTask(SoapObject rq, AsyncResponse delegate) {
    req1 = rq;
    this.delegate = delegate;
}

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

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

    SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11);
    envelope.setOutputSoapObject(this.req1);

    AndroidHttpTransport transport = new AndroidHttpTransport(Strings.URL_TSMS);
    transport.debug = true;

    String result = null;
    try {
        transport.call(Strings.URL_TSMS + this.req1.getName(), envelope);
        result = envelope.getResponse().toString();
    } catch (IOException ex) {
        Log.e("" , ex.getMessage());
    } catch (XmlPullParserException ex) {
        Log.e("" , ex.getMessage());
    }

    if (result != null && result.equals(String.valueOf(Integers.CODE_USER_PASS_FALSE))) {
        try {
            throw new TException(PublicErrorList.USERNAME_PASSWORD_ERROR);
        } catch (TException e) {
            e.printStackTrace();
        }

    }

    Log.e("------------++++++++++++++++-----------", result);

    return result;
}

protected void onPostExecute(String result) {
/* super.onPostExecute(result);*/
    delegate.processFinish(result);
}

}

现在您可以像这样从外部执行 ProcessTask，这将确保您异步接收结果:

new ProcessTask(rq, new AsyncResponse() {
    @Override
    public void processFinish(String output) {
        // do whatever you want with the result
    }
}).execute();