Jquery ajax调用没有命中servlet

I am trying to make a simple ajax call. No matter what I do, it always executes the error block. I have a sysout in the doPost that is never hit. Someone please tell me what I am doing wrong. Here is my code.

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中有一个从未被命中的sysout。有人请告诉我我做错了什么。这是我的代码。

javascript----

JavaScript的----

$.ajax({
    url: "GetBulletAjax",
    dataType: 'json',
    success: function(data) {
        alert("success");
    },
     error: function(jqXHR, textStatus, errorThrown) {
        alert(jqXHR+" - "+textStatus+" - "+errorThrown);
    }       
});

Java----

Java的----

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();
        }       
    }

}

web.xml----

web.xml中----

<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>

2 个解决方案

#1

What's the URL for your client? Your URL is going to be relative -- so if your page's URL is <server>/foo/bar.html, your ajax request is going to go to <server>/foo/GetBulletAjax. But your servlet definition is <server>/GetBulletAjax.

您客户的URL是什么？您的URL将是相对的 - 因此，如果您的页面的URL是 /foo/bar.html，那么您的ajax请求将转到 / foo / GetBulletAjax。但是你的servlet定义是 / GetBulletAjax。

Change your url in your ajax request to /GetBulletAjax. You need the leading forward slash to tell the browser the resource is located off the root of the site.

将您的ajax请求中的网址更改为/ GetBulletAjax。您需要前导斜杠来告诉浏览器资源位于站点根目录之外。

#2

in Jquery documentation

在Jquery文档中

http://api.jquery.com/jQuery.ajax/

type (default: 'GET') Type: String The type of request to make ("POST" or "GET"), default is "GET". Note: Other HTTP request methods, such as PUT and DELETE, can also be used here, but they are not supported by all browsers.

type（默认值：'GET'）类型：String要生成的请求类型（“POST”或“GET”），默认为“GET”。注意：此处也可以使用其他HTTP请求方法，例如PUT和DELETE，但并非所有浏览器都支持它们。

seems that you miss the type attribute which needs to be POST. default is GET as mentioned by documentation. You dont have a doGet in your servlet to support that.

好像你错过了需要POST的type属性。默认是文档中提到的GET。你的servlet中没有doGet来支持它。

$.ajax({
   url: "GetBulletAjax",
   dataType: 'json',
   type:POST,
   success: function(data) {
      alert("success");
   },
   error: function(jqXHR, textStatus, errorThrown) {
      alert(jqXHR+" - "+textStatus+" - "+errorThrown);
   }       
});

#1