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JavaScript json Ajax Servlet prototype

原文来自: http://wangyu.iteye.com/blog/311274

JSON （JavaScript Object Notation）一种简单的数据格式，比xml更轻巧。 JSON 是 JavaScript 原生格式，这意味着在 JavaScript 中处理 JSON 数据不需要任何特殊的 API 或工具包。
JSON的规则很简单：对象是一个无序的“‘名称/值’对”集合。一个对象以“{”（左括号）开始，“}”（右括号）结束。每个“名称”后跟一个“:”（冒号）；“‘名称/值’ 对”之间使用“,”（逗号）分隔。具体细节参考http://www.json.org/json-zh.html
举个简单的例子：
Js代码

   1. function showJSON() {
   2.     var user =
   3.     {
   4.         "username":"andy",
   5.         "age":20,
   6.         "info": { "tel": "123456", "cellphone": "98765"},
   7.         "address":
   8.             [
   9.                 {"city":"beijing","postcode":"222333"},
10.                 {"city":"newyork","postcode":"555666"}
11.             ]
12.     }
13.
14.     alert(user.username);
15.     alert(user.age);
16.     alert(user.info.cellphone);
17.     alert(user.address[0].city);
18.     alert(user.address[0].postcode);
19. }

function showJSON() {
    var user =
    {
        "username":"andy",
        "age":20,
        "info": { "tel": "123456", "cellphone": "98765"},
        "address":
            [
                {"city":"beijing","postcode":"222333"},
                {"city":"newyork","postcode":"555666"}
            ]
    }

    alert(user.username);
    alert(user.age);
    alert(user.info.cellphone);
    alert(user.address[0].city);
    alert(user.address[0].postcode);
}

这表示一个user对象，拥有username, age, info, address 等属性。
同样也可以用JSON来简单的修改数据，修改上面的例子
Js代码

   1. function showJSON() {
   2.     var user =
   3.     {
   4.         "username":"andy",
   5.         "age":20,
   6.         "info": { "tel": "123456", "cellphone": "98765"},
   7.         "address":
   8.             [
   9.                 {"city":"beijing","postcode":"222333"},
10.                 {"city":"newyork","postcode":"555666"}
11.             ]
12.     }
13.
14.     alert(user.username);
15.     alert(user.age);
16.     alert(user.info.cellphone);
17.     alert(user.address[0].city);
18.     alert(user.address[0].postcode);
19.
20.     user.username = "Tom";
21.     alert(user.username);
22. }

function showJSON() {
    var user =
    {
        "username":"andy",
        "age":20,
        "info": { "tel": "123456", "cellphone": "98765"},
        "address":
            [
                {"city":"beijing","postcode":"222333"},
                {"city":"newyork","postcode":"555666"}
            ]
    }

    alert(user.username);
    alert(user.age);
    alert(user.info.cellphone);
    alert(user.address[0].city);
    alert(user.address[0].postcode);

    user.username = "Tom";
    alert(user.username);
}

JSON提供了json.js包，下载[url]http://www.json.org/json.js [/url]后，将其引入然后就可以简单的使用object.toJSONString()转换成JSON数据。
Js代码

   1. function showCar() {
   2.     var carr = new Car("Dodge", "Coronet R/T", 1968, "yellow");
   3.     alert(carr.toJSONString());
   4. }
   5.
   6. function Car(make, model, year, color)       {
   7.      this.make = make;
   8.      this.model = model;
   9.      this.year = year;
10.      this.color = color;
11. }

function showCar() {
    var carr = new Car("Dodge", "Coronet R/T", 1968, "yellow");
    alert(carr.toJSONString());
}

function Car(make, model, year, color)       {
     this.make = make;
     this.model = model;
     this.year = year;
     this.color = color;
}

可以使用eval来转换JSON字符到Object
Js代码

   1. function myEval() {
   2.     var str = '{ "name": "Violet", "occupation": "character" }';
   3.     var obj = eval('(' + str + ')');
   4.     alert(obj.toJSONString());
   5. }

function myEval() {
    var str = '{ "name": "Violet", "occupation": "character" }';
    var obj = eval('(' + str + ')');
    alert(obj.toJSONString());
}

或者使用parseJSON()方法
Js代码

   1. function myEval() {
   2.     var str = '{ "name": "Violet", "occupation": "character" }';
   3.     var obj = str.parseJSON();
   4.     alert(obj.toJSONString());
   5. }

function myEval() {
    var str = '{ "name": "Violet", "occupation": "character" }';
    var obj = str.parseJSON();
    alert(obj.toJSONString());
}

下面使用prototype写一个JSON的ajax例子。
先写一个servlet (我的是servlet.ajax.JSONTest1.java)就写一句话
Java 代码

   1. response.getWriter().print("{ \"name\": \"Violet\", \"occupation\": \"character\" }");

response.getWriter().print("{ \"name\": \"Violet\", \"occupation\": \"character\" }");

再在页面中写一个ajax的请求
Js代码

   1. function sendRequest() {
   2.     var url = "/MyWebApp/JSONTest1";
   3.     var mailAjax = new Ajax.Request(
   4.         url,
   5.         {
   6.             method: 'get',
   7.             onComplete: jsonResponse
   8.         }
   9.     );
10. }
11.
12. function jsonResponse(originalRequest) {
13.     alert(originalRequest.responseText);
14.     var myobj = originalRequest.responseText.parseJSON();
15.     alert(myobj.name);
16. }

function sendRequest() {
    var url = "/MyWebApp/JSONTest1";
    var mailAjax = new Ajax.Request(
        url,
        {
            method: 'get',
            onComplete: jsonResponse
        }
    );
}

function jsonResponse(originalRequest) {
    alert(originalRequest.responseText);
    var myobj = originalRequest.responseText.parseJSON();
    alert(myobj.name);
}

prototype-1.5.1.js中提供了JSON的方法，String.evalJSON(), 可以不使用json.js, 修改上面的方法
Js代码

   1. function jsonResponse(originalRequest) {
   2.     alert(originalRequest.responseText);
   3.     var myobj = originalRequest.responseText.evalJSON(true);
   4.     alert(myobj.name);
   5. }

function jsonResponse(originalRequest) {
    alert(originalRequest.responseText);
    var myobj = originalRequest.responseText.evalJSON(true);
    alert(myobj.name);
}

JSON还提供了java的jar包 http://www.json.org/java/index.html API也很简单，下面举个例子
在javascript中填加请求参数
Js代码

   1. function sendRequest() {
   2.     var carr = new Car("Dodge", "Coronet R/T", 1968, "yellow");
   3.     var pars = "car=" + carr.toJSONString();
   4.
   5.     var url = "/MyWebApp/JSONTest1";
   6.     var mailAjax = new Ajax.Request(
   7.         url,
   8.         {
   9.             method: 'get',
10.             parameters: pars,
11.             onComplete: jsonResponse
12.         }
13.     );
14. }

function sendRequest() {
    var carr = new Car("Dodge", "Coronet R/T", 1968, "yellow");
    var pars = "car=" + carr.toJSONString();

    var url = "/MyWebApp/JSONTest1";
    var mailAjax = new Ajax.Request(
        url,
        {
            method: 'get',
            parameters: pars,
            onComplete: jsonResponse
        }
    );
}

使用JSON请求字符串就可以简单的生成JSONObject并进行解析,修改servlet添加JSON的处理(要使用json.jar)
Java 代码

   1. private void doService(HttpServletRequest request, HttpServletResponse response) throws IOException {
   2.         String s3 = request.getParameter("car");
   3.         try {
   4.             JSONObject jsonObj = new JSONObject(s3);
   5.             System.out.println(jsonObj.getString("model"));
   6.             System.out.println(jsonObj.getInt("year"));
   7.         } catch (JSONException e) {
   8.             e.printStackTrace();
   9.         }
10.         response.getWriter().print("{ \"name\": \"Violet\", \"occupation\": \"character\" }");
11.     }

private void doService(HttpServletRequest request, HttpServletResponse response) throws IOException {
        String s3 = request.getParameter("car");
        try {
            JSONObject jsonObj = new JSONObject(s3);
            System.out.println(jsonObj.getString("model"));
            System.out.println(jsonObj.getInt("year"));
        } catch (JSONException e) {
            e.printStackTrace();
        }
        response.getWriter().print("{ \"name\": \"Violet\", \"occupation\": \"character\" }");
    }

同样可以使用JSONObject生成JSON字符串，修改servlet
Java 代码

   1. private void doService(HttpServletRequest request, HttpServletResponse response) throws IOException {
   2.         String s3 = request.getParameter("car");
   3.         try {
   4.             JSONObject jsonObj = new JSONObject(s3);
   5.             System.out.println(jsonObj.getString("model"));
   6.             System.out.println(jsonObj.getInt("year"));
   7.         } catch (JSONException e) {
   8.             e.printStackTrace();
   9.         }
10.
11.         JSONObject resultJSON = new JSONObject();
12.         try {
13.             resultJSON.append("name", "Violet")
14.                       .append("occupation", "developer")
15.                       .append("age", new Integer(22));
16.             System.out.println(resultJSON.toString());
17.         } catch (JSONException e) {
18.             e.printStackTrace();
19.         }
20. }

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