这是我的jQuery代码我正在发出一个AJAX请求并将param发送给servlet。我可以看到参数的URL,但我没能获得请求参数在servlet
//jquery
<script>
$.post("dashboardcon", $("#date").serialize(), function(responseHtml) {
$('#date').html(responseHtml);
});
</script>
HTML
<form>
Date :<br> <input id="date" type="text" name="date">
<input type="submit" value="Submit">
</form>
的servlet
@WebServlet("/dashboardcon")
public class dashboardCon extends HttpServlet {
private static final long serialVersionUID = 1L;
protected void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
JSONObject jObj = new JSONObject();
try {
JSONObject newObj = jObj.getJSONObject(request.getParameter("date"));
System.out.println(newObj);
} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
您确定要发送到您servlet的正确URL吗? – Difster
是的!我检查了它 –
什么是后端语言,你如何检查那里的参数? – Difster