我有一个处理帖子和获取请求的代码。但它应该在发布请求时响应。我不想使用servlet因为它需要tomcat或jetty而且它变得更加复杂。 我应该怎么知道收到的帖子请求?
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());
}
public static void main(String[] args) throws Exception {
while (true) {
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();
}
}