我得到一个程序，从一个链接的服务器获取天气。我已经做了一些将字符串结合到URL的字符串。我现在需要从XML文件中获取信息。android：如何从xml文件中获取信息？

这是我的代码：(我更换了，为了安全起见API KEY)

package com.example.thelexapp;

import android.location.Location;

import android.location.LocationManager;

import android.os.Bundle;

import android.app.Activity;

import android.content.Context;

import android.graphics.Color;

import android.view.Menu;

import android.view.View;

import android.webkit.WebView;

import android.widget.TextView;

public class MainActivity extends Activity {

@Override

protected void onCreate(Bundle savedInstanceState) {

super.onCreate(savedInstanceState);

setContentView(R.layout.activity_main);

}

@Override

public boolean onCreateOptionsMenu(Menu menu) {

getMenuInflater().inflate(R.menu.main, menu);

return true;

}

public void getPositionAndGetWheater(View view) {

LocationManager lm = (LocationManager) getSystemService(Context.LOCATION_SERVICE);

Location location = lm.getLastKnownLocation(LocationManager.GPS_PROVIDER);

if (location != null) {

TextView text = (TextView) findViewById(R.id.textView1);

text.setTextColor(Color.parseColor("#808080"));

double longitude = location.getLongitude();

double latitude = location.getLatitude();

String longitudestring = String.valueOf(longitude);

String latitudestring = String.valueOf(latitude);

String URLforweather = "http://api.worldweatheronline.com/free/v1/weather.ashx?q=" + longitudestring + "," + latitudestring + "&format=xml&num_of_days=1&key=APIKEYIREPLACEDb";

WebView webview = (WebView) findViewById(R.id.webview);

webview.loadUrl(URLforweather);

} else {

final TextView textViewToChange = (TextView) findViewById(R.id.textView1);

textViewToChange.setText("Unable to get weather information.");

TextView text = (TextView) findViewById(R.id.textView1);

text.setTextColor(Color.parseColor("#FF0000"));

}

}

}