我已经设置了一个java servlet,它接受URL中的参数并使其正常工作:
public class GetThem extends HttpServlet {
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws IOException, ServletException
{
try {
double lat=Double.parseDouble(request.getParameter("lat"));
double lon=Double.parseDouble(request.getParameter("lon"));
response.setContentType("text/html");
PrintWriter out = response.getWriter();
out.println(lat + " and " + lon);
} catch (Exception e) {
e.printStackTrace();
}
}
}
因此,访问此链接:http://www.example.com:8080/HttpPost/HttpPost?lat=1&lon=2将输出:
"1.0 and 2.0"
我目前正在使用以下代码从另一个java程序调用它:
try{
URL objectGet = new URL("http://www.example.com:8080/HttpPost/HttpPost?lat=" + Double.toString(dg.getLatDouble()) + "&lon=" + Double.toString(dg.getLonDouble()));
URLConnection yc = objectGet.openConnection();
BufferedReader in = new BufferedReader(
new InputStreamReader(
yc.getInputStream()));
in = new BufferedReader(
new InputStreamReader(
yc.getInputStream()));
...
现在我想更改它,这样我就不会使用URL参数将这些数据传递到服务器。我想向这个服务器发送更大的消息。我知道我需要使用http post而不是http get来实现这一点,但不确定如何做到
我需要更改正在接收数据的服务器端的任何内容吗?我需要在发布这些数据的客户端做什么?
如有任何帮助,我们将不胜感激。理想情况下,我希望以JSON格式发送这些数据。
下面是"java HTTP POST example
"在谷歌中找到的第一个链接的示例。
try {
// Construct data
StringBuilder dataBuilder = new StringBuilder();
dataBuilder.append(URLEncoder.encode("key1", "UTF-8")).append('=').append(URLEncoder.encode("value1", "UTF-8")).
append(URLEncoder.encode("key2", "UTF-8")).append('=').append(URLEncoder.encode("value2", "UTF-8"));
// Send data
URL url = new URL("http://hostname:80/cgi");
URLConnection conn = url.openConnection();
conn.setDoOutput(true);
OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
wr.write(dataBuilder.toString());
wr.flush();
// Get the response
BufferedReader rd = new BufferedReader(new InputStreamReader(conn.getInputStream()));
String line;
while ((line = rd.readLine()) != null) {
// Process line...
}
wr.close();
rd.close();
} catch (Exception e) {
}
我认为应该使用HTTPClient,而不是处理连接和流。检查http://hc.apache.org/httpclient-3.x/tutorial.html