JSON格式解析 .
其中 Value 也可以是另一个 Object 或者数组,因此,复杂的 Object 可以嵌套表示,例如,一个 Person 对象包含 name 和 address 对象,可以表示如下:
{"name":"Michael","address":
??? {"city":"Beijing","street":" Chaoyang Road ","postcode":100025}
}
我们通过一个简单的 JavaScript 方法就能看到客户端如何将 JSON 数据表示给用户:
function handleJson() {
? var j={"name":"Michael","address":
????? {"city":"Beijing","street":" Chaoyang Road ","postcode":100025}
? };
? document.write(j.name);
? document.write(j.address.city);
?}
假定服务器返回的 JSON 数据是上文的:
{"name":"Michael","address":
??? {"city":"Beijing","street":" Chaoyang Road ","postcode":100025}
}
只需将其赋值给一个 JavaScript 变量,就可以立刻使用该变量并更新页面中的信息了
服务器端输出 JSON 格式数据
以 Java 为例,我们将演示将一个 Java 对象编码为 JSON 格式的文本。
将 String 对象编码为 JSON 格式时,只需处理好特殊字符即可。另外,必须用 (") 而非 (') 表示字符串:
?
?static String string2Json(String s) {
??? StringBuilder sb = new StringBuilder(s.length()+20);
??? sb.append('/"');
??? for (int i=0; i<s.length(); i++) {
??????? char c = s.charAt(i);
??????? switch (c) {
??????? case '/"':
??????????? sb.append("///"");
??????????? break;
??????? case '//':
??????????? sb.append("////");
??????????? break;
??????? case '/':
??????????? sb.append("///");
??????????? break;
??????? case '/b':
??????????? sb.append("//b");
??????????? break;
??????? case '/f':
??????????? sb.append("//f");
??????????? break;
??????? case '/n':
??????????? sb.append("//n");
??????????? break;
??????? case '/r':
??????????? sb.append("//r");
??????????? break;
??????? case '/t':
??????????? sb.append("//t");
??????????? break;
??????? default:
??????????? sb.append(c);
??????? }
??? }
??? sb.append('/"');
??? return sb.toString();
?}
?
将 Number 表示为 JSON 就容易得多,利用 Java 的多态,我们可以处理 Integer,Long,Float 等多种 Number 格式:
?
?static String number2Json(Number number) {
??? return number.toString();
?}
?
Boolean 类型也可以直接通过 toString() 方法得到 JSON 的表示:
?
?static String boolean2Json(Boolean bool) {
??? return bool.toString();
?}
?
要将数组编码为 JSON 格式,可以通过循环将每一个元素编码出来:
?
?static String array2Json(Object[] array) {
??? if (array.length==0)
??????? return "[]";
??? StringBuilder sb = new StringBuilder(array.length << 4);
??? sb.append('[');
??? for (Object o : array) {
??????? sb.append(toJson(o));
??????? sb.append(',');
??? }
??? // 将最后添加的 ',' 变为 ']':
??? sb.setCharAt(sb.length()-1, ']');
??? return sb.toString();
?}
?
最后,我们需要将 Map<String, Object> 编码为 JSON 格式,因为 JavaScript 的 Object 实际上对应的是 Java 的 Map<String, Object> 。该方法如下:
?
?static String map2Json(Map<String, Object> map) {
??? if (map.isEmpty())
??????? return "{}";
??? StringBuilder sb = new StringBuilder(map.size() << 4);
??? sb.append('{');
??? Set<String> keys = map.keySet();
??? for (String key : keys) {
??????? Object value = map.get(key);
??????? sb.append('/"');
??????? sb.append(key);
??????? sb.append('/"');
??????? sb.append(':');
??????? sb.append(toJson(value));
??????? sb.append(',');
??? }
??? // 将最后的 ',' 变为 '}':
??? sb.setCharAt(sb.length()-1, '}');
??? return sb.toString();
?}
为了统一处理任意的 Java 对象,我们编写一个入口方法 toJson(Object),能够将任意的 Java 对象编码为 JSON 格式:
?
?public static String toJson(Object o) {
??? if (o==null)
??????? return "null";
??? if (o instanceof String)
??????? return string2Json((String)o);
??? if (o instanceof Boolean)
??????? return boolean2Json((Boolean)o);
??? if (o instanceof Number)
??????? return number2Json((Number)o);
??? if (o instanceof Map)
??????? return map2Json((Map<String, Object>)o);
??? if (o instanceof Object[])
??????? return array2Json((Object[])o);
??? throw new RuntimeException("Unsupported type: " + o.getClass().getName());
?}
?
);??再在页面中写一个ajax的请求
js 代码- 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- 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- 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- privatevoid 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- privatevoid 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(); ?????????? } ?????????? ??????????JSONObject resultJSON = new JSONObject(); ??????????try { ?????????????? resultJSON.append("name", "Violet") ???????????????????????? .append("occupation", "developer") ???????????????????????? .append("age", new Integer(22)); ?????????????? System.out.println(resultJSON.toString()); ?????????? } catch (JSONException e) { ?????????????? e.printStackTrace(); ?????????? } ?????????? response.getWriter().print(resultJSON.toString()); ?????? } ??
- function jsonResponse(originalRequest) { ?????? alert(originalRequest.responseText); ??????var myobj = originalRequest.responseText.evalJSON(true); ?????? alert(myobj.name); ?????? alert(myobj.age); ??}??
参考
http://www.json.org/js.html
http://www.blogjava.net/Jkallen/archive/2006/03/28/37905.html
http://www.json.org/
http://www.prototypejs.org/learn/json
http://www.json.org/java/index.html
http://www.ibm.com/developerworks/cn/web/wa-ajaxintro10/index.html
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