我想要做的就是使用AsyncTask来执行http请求 . 这是我的代码到目前为止，但我不知道如何从我的主要活动中调用此类以及如何获得结果 . 我有一个String var“uri”，它完成了我的http请求的url，我必须传递给我的类 . 在调用此类之后，下一步是将结果(作为字符串)传递给新函数 .

import java.io.ByteArrayOutputStream;

import java.io.IOException;

import org.apache.http.HttpResponse;

import org.apache.http.HttpStatus;

import org.apache.http.StatusLine;

import org.apache.http.client.ClientProtocolException;

import org.apache.http.client.HttpClient;

import org.apache.http.client.methods.HttpGet;

import org.apache.http.impl.client.DefaultHttpClient;

import android.os.AsyncTask;

class RequestTask extends AsyncTask{

@Override

protected String doInBackground(String... uri) {

HttpClient httpclient = new DefaultHttpClient();

HttpResponse response;

String responseString = null;

try {

response = httpclient.execute(new HttpGet("http://www.mywebsite.com/xml_data/" + uri + "_meeting_info.xml?id=cacheSTOP3"));

StatusLine statusLine = response.getStatusLine();

if(statusLine.getStatusCode() == HttpStatus.SC_OK){

ByteArrayOutputStream out = new ByteArrayOutputStream();

response.getEntity().writeTo(out);

out.close();

responseString = out.toString();

} else{

//Closes the connection.

response.getEntity().getContent().close();

throw new IOException(statusLine.getReasonPhrase());

}

} catch (ClientProtocolException e) {

//TODO Handle problems..

} catch (IOException e) {

//TODO Handle problems..

}

return responseString;

}

@Override

protected void onPostExecute(String result) {

onPostExecute(result);

//Do anything with response..

}

}

主要活动：

RequestTask getXML = new RequestTask();

getXML.execute(MyVarString);

//Now send the results from above to a new function

Document doc = XMLfunctions.XMLfromString(getdata);