如何(完全)将查询字符串转换为JSON对象？

Question

如果我有以下表格:

<form>
  <input name="foo" value="bar">
  <input name="hello" value="world">
  <input name="animals[]" value="panda">
  <input name="animals[]" value="koala">
  <input name="car[make]" value="Honda">
  <input name="car[origin]" value="Japan">
</form>

我不希望使用$("form").serialize():

foo=bar&hello=world&animals%5B%5D=panda&animals%5B%5D=koalacar&%5Bmake%5D=Honda&car%5Borigin%5D=Japan

相反,我想要这个:

{"foo":"bar", "hello":"world", "animals":["panda", "koala"], "car":{"make":"Honda", "origin":"Japan"}}

据我所知,jQuery过去常常这样做,但他们切换serialize方法返回GET风格的查询字符串.有没有简单的方法来获得我想要的结果？

---

编辑

我已将原始问题更新为包含car[make]和car[origin]示例.应该假设foo[bar][baz]或foo[bar][baz][bof]输入也可以出现在表格上.

另外,被诸如指定数字索引键foo[0]=a,foo[1]=b,foo[4]=c应该被保留,例如

{ ... "foo":["a", "b", undefined, undefined, "c"] ... }

Answer 1

将表单转换为JSON LIKE A BOSS

Github:跟随Github

以下代码可以使用各种输入名称; 并按照您的期望处理它们.

例如,

<!-- all of these will work! -->
<input name="honey[badger]" value="a">
<input name="wombat[]" value="b">
<input name="hello[panda][]" value="c">
<input name="animals[0][name]" value="d">
<input name="animals[0][breed]" value="e">
<input name="crazy[1][][wonky]" value="f">
<input name="dream[as][vividly][as][you][can]" value="g">

// output
{
  "honey":{
    "badger":"a"
  },
  "wombat":["b"],
  "hello":{
    "panda":["c"]
  },
  "animals":[
    {
      "name":"d",
      "breed":"e"
    }
  ],
  "crazy":[
    null,
    [
      {"wonky":"f"}
    ]
  ],
  "dream":{
    "as":{
      "vividly":{
        "as":{
          "you":{
            "can":"g"
          }
        }
      }
    }
  }
}

用法

$('#my-form').serializeObject();

巫术

(function($){
    $.fn.serializeObject = function(){

        var self = this,
            json = {},
            push_counters = {},
            patterns = {
                "validate": /^[a-zA-Z][a-zA-Z0-9_]*(?:\[(?:\d*|[a-zA-Z0-9_]+)\])*$/,
                "key":      /[a-zA-Z0-9_]+|(?=\[\])/g,
                "push":     /^$/,
                "fixed":    /^\d+$/,
                "named":    /^[a-zA-Z0-9_]+$/
            };


        this.build = function(base, key, value){
            base[key] = value;
            return base;
        };

        this.push_counter = function(key){
            if(push_counters[key] === undefined){
                push_counters[key] = 0;
            }
            return push_counters[key]++;
        };

        $.each($(this).serializeArray(), function(){

            // skip invalid keys
            if(!patterns.validate.test(this.name)){
                return;
            }

            var k,
                keys = this.name.match(patterns.key),
                merge = this.value,
                reverse_key = this.name;

            while((k = keys.pop()) !== undefined){

                // adjust reverse_key
                reverse_key = reverse_key.replace(new RegExp("\\[" + k + "\\]$"), '');

                // push
                if(k.match(patterns.push)){
                    merge = self.build([], self.push_counter(reverse_key), merge);
                }

                // fixed
                else if(k.match(patterns.fixed)){
                    merge = self.build([], k, merge);
                }

                // named
                else if(k.match(patterns.named)){
                    merge = self.build({}, k, merge);
                }
            }

            json = $.extend(true, json, merge);
        });

        return json;
    };
})(jQuery);

Answer 2

您可以使用".serializeArray()",然后修复结果:

var json = {};
$.each($('form').serializeArray(), function() {
  json[this.name] = this.value;
});

当然,您可能需要担心多值字段:

var json = {};
$.each($('form').serializeArray(), function() {
  var cur = json[this.name];
  if (cur !== undefined) {
    if ($.isArray(cur))
      cur.push(this.value);
    else
      json[ this.name.replace(/\[[^\]]*\]$/, '') ] = [ cur, this.value ];
  }
  else
    json[this.name] = this.value;
});

(编辑 - 现在我想起来了,"序列化"和"serializeArray"已经为你处理了多值参数,给你起了序列化形式的"无论[2]"这样的名字.无论如何它都会工作但除了简单的事情之外,可能没有必要做更多的事情.)