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

这里有一种非jQuery方法:

    var getFormData = function(form) {
        //Ignore the submit button
        var elements = Array.prototype.filter.call(form.elements, function(element) {
            var type = element.getAttribute('type');
            return !type || type.toLowerCase() !== 'submit';
        });

您可以这样使用:

function() {

    var getFormData = function(form) {
        //Ignore the submit button
        var elements = Array.prototype.filter.call(form.elements, function(element) {
            var type = element.getAttribute('type');
            return !type || type.toLowerCase() !== 'submit';
        });

        //Make an object out of the form data: {name: value}
        var data = elements.reduce(function(data, element) {
            data[element.name] = element.value;
            return data;
        }, {});

        return data;
    };

    var post = function(action, data, callback) {
        var request = new XMLHttpRequest();
        request.onload = callback;
        request.open('post', action);
        request.setRequestHeader("Content-Type", "application/json;charset=UTF-8");
        request.send(JSON.stringify(data), true);
        request.send();
    };

    var submit = function(e) {
        e.preventDefault();
        var form = e.target;
        var action = form.action;
        var data = getFormData(form);
        //change the third argument in order to do something
        //more intersting with the response than just print it
        post(action, data, console.log.bind(console));
    }

    //change formName below
    document.formName.onsubmit = submit;

})();

其他回答

这和你想要的完全一样

仅执行以下代码一次

$.fn.serializeObject = function(){
    let d={};
    $(this).serializeArray().forEach(r=>d[r.name]=r.value);
    return d;
}

现在您可以多次执行以下行

let formObj = $('#myForm').serializeObject();
// will return like {id:"1", username:"abc"}

使用maček的解决方案,我对其进行了修改,使其与ASP.NET MVC在同一表单上处理嵌套/复杂对象的方式一致。您所要做的就是将验证部分修改为:

"validate": /^[a-zA-Z][a-zA-Z0-9_]*((?:\[(?:\d*|[a-zA-Z0-9_]+)\])*(?:\.)[a-zA-Z][a-zA-Z0-9_]*)*$/,

这将匹配并正确映射具有以下名称的元素:

<input type="text" name="zooName" />

And

<input type="text" name="zooAnimals[0].name" />

此代码适用于我:

  var data = $('#myForm input, #myForm select, #myForm textarea').toArray().reduce(function (m, e) {
            m[e.name] = $(e).val();
            return m;
        }, {});

Use:

function form_to_json (selector) {
  var ary = $(selector).serializeArray();
  var obj = {};
  for (var a = 0; a < ary.length; a++) obj[ary[a].name] = ary[a].value;
  return obj;
}

输出:

{"myfield": "myfield value", "passwordfield": "mypasswordvalue"}

此线程似乎已成为表单序列化的常见问题解答:)

我对PHP命名的看法:<input name=“user[name]”>

$('form').on('submit', function(ev) {
   ev.preventDefault();

   var obj = $(this).serializePHPObject();

   // $.post('./', obj);
});
(function ($) {
  // based on https://stackoverflow.com/a/25239999/1644202

  // <input name="user[name]" >
  $.fn.serializePHPObject = function () {
    var obj = {};
    $.each(this.serializeArray(), function (i, pair) {
      var cObj = obj,
        pObj,
        cpName;
      $.each(pair.name.split("["), function (i, pName) {
        pName = pName.replace("]", "");
        pObj = cObj;
        cpName = pName;
        cObj = cObj[pName] ? cObj[pName] : (cObj[pName] = {});
      });
      pObj[cpName] = pair.value;
    });
    return obj;
  };
})(jQuery);