我想做一个简单的ajax调用。无论我做什么,它总是执行错误块。我在doPost中有一个从未命中的sysout。有人请告诉我我做错了什么。这是我的代码。jquery ajax调用不是打到servlet
的JavaScript ----
$.ajax({
url: "GetBulletAjax",
dataType: 'json',
success: function(data) {
alert("success");
},
error: function(jqXHR, textStatus, errorThrown) {
alert(jqXHR+" - "+textStatus+" - "+errorThrown);
}
});
的Java ----
public class GetBulletAjax extends HttpServlet {
private static final long serialVersionUID = 1L;
public GetBulletAjax() {
super();
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
doPost(request, response);
}
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
System.out.println("made it to servlet");
PrintWriter out = response.getWriter();
User user = (User) request.getSession().getAttribute("user");
int userId = user.getId();
List<Bullet> bullets;
BulletDAO bulletdao = new BulletDAOImpl();
try {
bullets = bulletdao.findBulletsByUser(userId);
Gson gson = new Gson();
String json = gson.toJson(bullets);
System.out.println(json);
out.println(json);
out.close();
} catch (SQLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
的web.xml ----
<servlet>
<servlet-name>GetBulletAjax</servlet-name>
<servlet-class>bulletAjax.GetBulletAjax</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>GetBulletAjax</servlet-name>
<url-pattern>/GetBulletAjax</url-pattern>
</servlet-mapping>
你回来的错误是什么?什么是'textStatus'和'errorThrown'? – 2013-04-29 00:15:29
警报显示“[object Object] - error - Not Found” – Justin 2013-04-29 00:57:01