Question

嗯，我完全没有这个，我有所有代码来解析xml的关注和响应，以及发送它的网址，但我不知道如何发送它以及如何收到回复。我找不到任何完整的指南或其他东西。提前谢谢，抱歉我的英语不好。

Answer 1

public class HttpURLConnectionExample {

    private final String USER_AGENT = "Mozilla/5.0";

    public static void main(String[] args) throws Exception {

        HttpURLConnectionExample http = new HttpURLConnectionExample();

        System.out.println("Testing 1 - Send Http GET request");
        http.sendGet();

        System.out.println("\nTesting 2 - Send Http POST request");
        http.sendPost();

    }

    // HTTP GET request
    private void sendGet() throws Exception {

        String url = "http://www.google.com/search?q=mkyong";

        URL obj = new URL(url);
        HttpURLConnection con = (HttpURLConnection) obj.openConnection();

        // optional default is GET
        con.setRequestMethod("GET");

        //add request header
        con.setRequestProperty("User-Agent", USER_AGENT);

        int responseCode = con.getResponseCode();
        System.out.println("\nSending 'GET' request to URL : " + url);
        System.out.println("Response Code : " + responseCode);

        BufferedReader in = new BufferedReader(
                new InputStreamReader(con.getInputStream()));
        String inputLine;
        StringBuffer response = new StringBuffer();

        while ((inputLine = in.readLine()) != null) {
            response.append(inputLine);
        }
        in.close();

        //print result
        System.out.println(response.toString());

    }

    // HTTP POST request
    private void sendPost() throws Exception {

        String url = "https://selfsolve.apple.com/wcResults.do";
        URL obj = new URL(url);
        HttpsURLConnection con = (HttpsURLConnection) obj.openConnection();

        //add reuqest header
        con.setRequestMethod("POST");
        con.setRequestProperty("User-Agent", USER_AGENT);
        con.setRequestProperty("Accept-Language", "en-US,en;q=0.5");

        String urlParameters = "sn=C02G8416DRJM&cn=&locale=&caller=&num=12345";

        // Send post request
        con.setDoOutput(true);
        DataOutputStream wr = new DataOutputStream(con.getOutputStream());
        wr.writeBytes(urlParameters);
        wr.flush();
        wr.close();

        int responseCode = con.getResponseCode();
        System.out.println("\nSending 'POST' request to URL : " + url);
        System.out.println("Post parameters : " + urlParameters);
        System.out.println("Response Code : " + responseCode);

        BufferedReader in = new BufferedReader(
                new InputStreamReader(con.getInputStream()));
        String inputLine;
        StringBuffer response = new StringBuffer();

        while ((inputLine = in.readLine()) != null) {
            response.append(inputLine);
        }
        in.close();

        //print result
        System.out.println(response.toString());

    }

}

Answer 2

最佳解决方案是使用SOAPConnectionFactory

SOAPConnectionFactory soapConnectionFactory =  SOAPConnectionFactory.newInstance();
SOAPConnection connection = soapConnectionFactory.createConnection();
java.net.URL endpoint = new URL("url");
SOAPMessage message = xmlStringParser.getSoapMessageFromString(XMLString);
SOAPMessage response = connection.call(message, endpoint);

感谢大家！：d

如何在java中使用webservice？

2 个答案: