如何从AsyncTask获取字符串？

Question

我有以下课程:

public class getURLData extends AsyncTask<String, Integer, String>{

@Override
protected String doInBackground(String... params) {
    String line;
    try {  
        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(params[0]);

        HttpResponse httpResponse = httpClient.execute(httpPost);
        HttpEntity httpEntity = httpResponse.getEntity();
        line = EntityUtils.toString(httpEntity);

    } catch (UnsupportedEncodingException e) {
        line = "<results status=\"error\"><msg>Can't connect to server</msg></results>";
    } catch (MalformedURLException e) {
        line = "<results status=\"error\"><msg>Can't connect to server</msg></results>";
    } catch (IOException e) {
        line = "<results status=\"error\"><msg>Can't connect to server</msg></results>";
    }
    return line;
}

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

}

而我试图这样称呼它:

String output = null;
output = new getURLData().execute("http://www.domain.com/call.php?locationSearched=" + locationSearched);

但输出变量没有获取数据,而是我收到错误:

Type mismatch: cannot convert from AsyncTask<String,Integer,String> to String

Answer 1

该方法execute返回AynscTask自身,需要调用get:

output =
    new getURLData()
        .execute("http://www.example.com/call.php?locationSearched=" + locationSearched)
        .get();

这将启动一个新线程(via execute),同时阻塞当前线程(via get),直到新线程的工作完成并返回结果.

如果这样做,您只需将异步任务转换为同步任务即可.

但是,使用的问题get是因为它阻塞,所以需要在工作线程上调用它.但是,AsyncTask.execute()需要在主线程上调用.因此,尽管此代码可行,但您可能会得到一些不良结果.我还怀疑get()谷歌未经测试,他们可能会在某个地方引入一个错误.

参考:AsyncTask.get

Answer 2

我宁愿创建回调而不是阻止UI线程.在您的类创建方法,将在数据到达时调用.例如:

private void setData(String data){
    mTextView.setText(data);
}

然后在AsyncTask中实现onPostExecute:

@Override
protected void onPostExecute(String result) {
    setData(result);
}

然后在代码中的某处执行任务:

new getURLData().execute(...

当任务完成时,将调用setData并填充mTextView.

AsyncTask.get()将填充您的UI,因此没有理由使用AsyncTask.