在的Java Servlet通过jQuery AJAX GET参数发送 [英] Get parameter sent via jquery ajax in Java Servlet

问题描述

我搜索网络这个话题，但我不能让那个工作为例。 我将gladed与有人可以给我一个帮助。

I search this topic on web but I can't get a example that worked. I will be gladed with someone could give me a help.

这是我的测试。

 $.ajax({
    url: 'GetJson',
    type: 'POST',        
    dataType: 'json',
    contentType: 'application/json',

    data: {id: 'idTest'},
    success: function(data) {
        console.log(data);
    }
});

在Sevlet

protected void doPost(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {

    String id = request.getParameter("id");
    String id2[] = request.getParameterValues("id");        
    String id3 = request.getHeader("id");

}

我收到空的一切。

I'm getting null in everything.

推荐答案

排序的答案是，这个数据被隐藏在请求的InputStream 。

The sort answer is that this data is hidden in the request InputStream.

下面的servlet是如何使用该演示（我在的JBoss 7.1.1运行它）：

The following servlet is a demo of how you can use this (I am running it on a JBoss 7.1.1):

import java.io.ByteArrayOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.UnsupportedEncodingException;
import java.net.URLDecoder;
import java.util.Enumeration;
import java.util.HashMap;
import java.util.Map;

import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

@WebServlet(name="fooServlet", urlPatterns="/foo")
public class FooServlet extends HttpServlet
{
    @Override
    protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
        InputStream is = req.getInputStream();
        ByteArrayOutputStream os = new ByteArrayOutputStream();
        byte[] buf = new byte[32];
        int r=0;
        while( r >= 0 ) {
            r = is.read(buf);
            if( r >= 0 ) os.write(buf, 0, r);
        }
        String s = new String(os.toByteArray(), "UTF-8");
        String decoded = URLDecoder.decode(s, "UTF-8");
        System.err.println(">>>>>>>>>>>>> DECODED: " + decoded);

        System.err.println("================================");

        Enumeration<String> e = req.getParameterNames();
        while( e.hasMoreElements() ) {
            String ss = (String) e.nextElement();
            System.err.println("    >>>>>>>>> " + ss);
        }

        System.err.println("================================");

        Map<String,String> map = makeQueryMap(s);
        System.err.println(map);
        //////////////////////////////////////////////////////////////////
        //// HERE YOU CAN DO map.get("id") AND THE SENT VALUE WILL BE ////
        //// RETURNED AS EXPECTED WITH request.getParameter("id")     ////
        //////////////////////////////////////////////////////////////////

        System.err.println("================================");

        resp.setContentType("application/json; charset=UTF-8");
        resp.getWriter().println("{'result':true}");
    }

    // Based on code from: http://www.coderanch.com/t/383310/java/java/parse-url-query-string-parameter
    private static Map<String, String> makeQueryMap(String query) throws UnsupportedEncodingException {
        String[] params = query.split("&");
        Map<String, String> map = new HashMap<String, String>();
        for( String param : params ) {
            String[] split = param.split("=");
            map.put(URLDecoder.decode(split[0], "UTF-8"), URLDecoder.decode(split[1], "UTF-8"));
        }
        return map;
    }
}

通过请求：

$.post("foo",{id:5,name:"Nikos",address:{city:"Athens"}})

的输出是：

>>>>>>>>>>>>> DECODED: id=5&name=Nikos&address[city]=Athens
================================
================================
{address[city]=Athens, id=5, name=Nikos}
================================

（注： req.getParameterNames（）不起作用印在4号线的地图包含了所有的数据通常可使用请求。 。的getParameter（）还要注意嵌套对象表示法， {地址：{城市：雅典}} ＆RARR; 地址[城市] =雅典）

(NOTE: req.getParameterNames() does not work. The map printed in the 4th line contains all the data normally accessible using request.getParameter(). Note also the nested object notation, {address:{city:"Athens"}} → address[city]=Athens)

略无关的问题，但出于完整性的：

Slightly unrelated to your question, but for the sake of completeness:

如果你想使用一个服务器端的JSON解析器，你应该使用 JSON.stringify 的数据：

If you want to use a server-side JSON parser, you should use JSON.stringify for the data:

$.post("foo",JSON.stringify({id:5,name:"Nikos",address:{city:"Athens"}}))

在我看来，与服务器进行通信JSON的最好方法是使用JAX-RS（或者Spring同等学历）。它是现代服务器死了简单而解决这些问题。

In my opinion the best way to communicate JSON with the server is using JAX-RS (or the Spring equivalent). It is dead simple on modern servers and solves these problems.

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