如何通过构造AJAX参数实现表单元素JSON相互转换？
ajax提交服务器数据， 整理一下转换方法。
HTML：
<form id="fm" name="fm" action=""> <input name="UserName" type="text" value="UserName1"/> </form> <input name="UserId" id="UserId" type="text" value="UserId1"/>
1.表单元素转QueryString
var q = $('#fm,#UserId').serialize(); //q = UserName=UserName1&UserId=UserId1
2.字符串， JSON 互相转换
var obj = jQuery.parseJSON('{"name":"John"}');
alert( obj.name === "John" );
可以利用jquery-json插件实现转换，直接引用示例
var thing = {plugin: 'jquery-json', version: 2.3};
var encoded = $.toJSON( thing );
// '{"plugin":"jquery-json","version":2.3}'
var name = $.evalJSON( encoded ).plugin;
// "jquery-json"
var version = $.evalJSON(encoded).version;
// 2.3
3.表单，元素转Name,Value数组
var arr = $("#fm,#UserId").serializeArray();
/*[
{name: 'UserName', value: '"UserName"1'},
{name: 'UserId', value: 'UserId'}
] */
4. 表单元素转JSON
$.fn.serializeObject = function()
{
var o = {};
var a = this.serializeArray();
$.each(a, function() {
if (o[this.name] !== undefined) {
if (!o[this.name].push) {
o[this.name] = [o[this.name]];
}
o[this.name].push(this.value || '');
} else {
o[this.name] = this.value || '';
}
});
return o;
};
var obj = $('form').serializeObject();
/*obj: Object
UserId: "UserId1"
UserName: "UserName1"
__proto__: Object*/
5. JSON2FORM
$.getJSON('url_to_file', function(data) {
for (var i in data) {
$('input[name="'+i+'"]').val(data[i]);
}
}
Google过程中发现一个更强大的插件 http://code.google.com/p/jquery-load-json/
data = {
"Name":"Emkay Entertainments",
"Address":"Nobel House, Regent Centre",
"Contact":"Phone"
}
$('div#data').loadJSON(data);
<div id="data">
<h1 id="Name">Emkay Entertainments</h1>
<label for="Address">Address:</label>
<span id="Address">Nobel House, Regent Centre</span>
<label for="Contact">Contact by:</label>
<span id="Contact">Phone</span>
</div>
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