如何将表单的所有元素转换为JavaScript对象?

我希望有某种方法从表单自动构建JavaScript对象,而不必遍历每个元素。我不希望使用$('#formid').serialize();返回的字符串;,我也不希望$('#formid').serializeArray()返回映射;


当前回答

const formData = new FormData(form);

let formDataJSON = {};

for (const [key, value] of formData.entries()) {

    formDataJSON[key] = value;
}

其他回答

要获得快速、现代的解决方案,请使用JSONify jQuery插件。下面的示例是从GitHub README中逐字摘录的。这一切都归功于插件作者Kushal Pandya。

鉴于:

<form id="myform">
    <label>Name:</label>
    <input type="text" name="name"/>
    <label>Email</label>
    <input type="text" name="email"/>
    <label>Password</label>
    <input type="password" name="password"/>
</form>

正在运行:

$('#myform').jsonify();

生产:

{"name":"Joe User","email":"joe@example.com","password":"mypass"}

如果要使用此JSON对象执行jQuery POST:

$('#mybutton').click(function() {
    $.post('/api/user', JSON.stringify($('#myform').jsonify()));
}

将任何东西变成对象(未经单元测试)

<script type="text/javascript">
string = {};

string.repeat = function(string, count)
{
    return new Array(count+1).join(string);
}

string.count = function(string)
{
    var count = 0;

    for (var i=1; i<arguments.length; i++)
    {
        var results = string.match(new RegExp(arguments[i], 'g'));
        count += results ? results.length : 0;
    }

    return count;
}

array = {};

array.merge = function(arr1, arr2)
{
    for (var i in arr2)
    {
        if (arr1[i] && typeof arr1[i] == 'object' && typeof arr2[i] == 'object')
            arr1[i] = array.merge(arr1[i], arr2[i]);
        else
            arr1[i] = arr2[i]
    }

    return arr1;
}

array.print = function(obj)
{
    var arr = [];
    $.each(obj, function(key, val) {
        var next = key + ": ";
        next += $.isPlainObject(val) ? array.print(val) : val;
        arr.push( next );
      });

    return "{ " +  arr.join(", ") + " }";
}

node = {};

node.objectify = function(node, params)
{
    if (!params)
        params = {};

    if (!params.selector)
        params.selector = "*";

    if (!params.key)
        params.key = "name";

    if (!params.value)
        params.value = "value";

    var o = {};
    var indexes = {};

    $(node).find(params.selector+"["+params.key+"]").each(function()
    {
        var name = $(this).attr(params.key),
            value = $(this).attr(params.value);

        var obj = $.parseJSON("{"+name.replace(/([^\[]*)/, function()
        {
            return '"'+arguments[1]+'"';
        }).replace(/\[(.*?)\]/gi, function()
        {
            if (arguments[1].length == 0)
            {
                var index = arguments[3].substring(0, arguments[2]);
                indexes[index] = indexes[index] !== undefined ? indexes[index]+1 : 0;

                return ':{"'+indexes[index]+'"';
            }
            else
                return ':{"'+escape(arguments[1])+'"';
        })+':"'+value.replace(/[\\"]/gi, function()
        {
            return "\\"+arguments[0]; 
        })+'"'+string.repeat('}', string.count(name, ']'))+"}");

        o = array.merge(o, obj);
    });

    return o;
}
</script>

测试输出:

$(document).ready(function()
{
    console.log(array.print(node.objectify($("form"), {})));
    console.log(array.print(node.objectify($("form"), {selector: "select"})));
});

on

<form>
    <input name='input[a]' type='text' value='text'/>
    <select name='input[b]'>
        <option>select</option>
    </select>

    <input name='otherinput[c][a]' value='a'/>
    <input name='otherinput[c][]' value='b'/>
    <input name='otherinput[d][b]' value='c'/>
    <input name='otherinput[c][]' value='d'/>

    <input type='hidden' name='anotherinput' value='hidden'/>
    <input type='hidden' name='anotherinput' value='1'/>

    <input type='submit' value='submit'/>
</form>

将产生:

{ input: { a: text, b: select }, otherinput: { c: { a: a, 0: b, 1: d }, d: { b: c } }, anotherinput: 1 }
{ input: { b: select } }

此函数返回转换为正确类型的所有值;

bool/string/(integer/floats)可能

虽然您需要jQuery来实现这一点,但由于serializeArray也是jQuery,所以没什么大不了的。

/**
 * serialized a form to a json object
 *
 * @usage: $("#myform").jsonSerialize();
 *
 */

(function($) {
    "use strict";
    $.fn.jsonSerialize = function() {
        var json = {};
        var array = $(this).serializeArray();
        $.each(array, function(key, obj) {
            var value = (obj.value == "") ? false : obj.value;
            if(value) {
                // check if we have a number
                var isNum = /^\d+$/.test(value);
                if(isNum) value = parseFloat(value);
                // check if we have a boolean
                var isBool = /^(false|true)+$/.test(value);
                if(isBool) value = (value!=="false");
            }
            json[obj.name] = value;
        });
        return json;
    }
})(jQuery);

这种解决方案更好。这里的一些更流行的选项在未选中复选框时不会更正句柄复选框。

getData: function(element){
//@todo may need additional logic for radio buttons
var select = $(element).find('select');
var input = $(element).find('input');
var inputs = $.merge(select,input);
var data = {};
//console.log(input,'input');
$.each(inputs,function(){
  if($(this).attr('type') != undefined){
    switch($(this).attr('type')){
     case 'checkbox':
        data[$(this).attr('name')] = ( ($(this).attr('checked') == 'checked') ? $(this).val():0 );
        break;
      default:
        data[$(this).attr('name')] = $(this).val();
        break;
    }
  } 
  else {
    data[$(this).attr('name')] = $(this).val();
  }
})
  return data;
}

使用lodash#set

let serialized = [
  { key: 'data[model][id]', value: 1 },
  { key: 'data[model][name]', value: 'product' },
  { key: 'sid', value: 'dh0un1hr4d' }
];

serialized.reduce(function(res, item) {
  _.set(res, item.key, item.value);
  return res;
}, {});

// returns
{
  "data": {
    "model": {
      "id": 1,
      "name": "product"
    }
  },
  "sid": "dh0un1hr4d"
}