Question

我这样称呼：

new asyncTaskPost().execute();

这是我的AsyncTask：

class asyncTaskPost extends AsyncTask<String, Void, Void> {

@Override
protected void onPreExecute() {
}

@Override
protected Void doInBackground(String... params) { <------ HERE ERROR!!!!!!
postMethod();
}

@Override
protected void onPostExecute(Void result) {
}

}

那个POST请求：

public static void postMethod()
{

HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://iwindroids.ru/test.php");
try {
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);

nameValuePairs.add(new BasicNameValuePair("lal", "ss"));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

HttpResponse response= httpclient.execute(httppost);
HttpEntity responseEntity = response.getEntity();
String serverResponse = EntityUtils.toString(responseEntity);
Log.i("server response is="+serverResponse,"");

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

提示问题protected Void doInBackground(String... params) {以及如何解决问题？我在代码中标记了一个错误

＆lt; ------ HERE ERROR !!!!!!

谢谢！

Answer 1

没有任何内容传递给doInbackground，您还需要一个返回语句

@Override
protected Void doInBackground(Void... params) {
postMethod();
return null;
}

同时更改

class asyncTaskPost extends AsyncTask<Void, Void, Void> {

Params，执行时发送给任务的参数类型。

检查AsyncTask泛型类型@

http://developer.android.com/reference/android/os/AsyncTask.html

AsyncTask中的POST请求错误

1 个答案: