Android调用包含url中参数的XML Webservice

Question

我正在尝试使用包含 web 服务参数的 get 方法调用 web 服务。但我无法在互联网上找到答案，请任何人帮助我。下面给出我的网络服务

http://api.crmseries.com/user/ValidateUser?email=don@crmSerssies.com&password=visi

Answer 1

这应该有效!

public void postData() {
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost("http://api.crmseries.com/user/ValidateUser");

try {
    // Add your data
    List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
    nameValuePairs.add(new BasicNameValuePair("email", "don@crmSerssies.com"));
    nameValuePairs.add(new BasicNameValuePair("password", "visi"));
    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

    // Execute HTTP Post Request
    HttpResponse response = httpclient.execute(httppost);

} catch (ClientProtocolException e) {
    // TODO Auto-generated catch block
} catch (IOException e) {
    // TODO Auto-generated catch block
}
}