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;
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
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);
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;
}, {});
所有这些答案对我来说似乎都太过头了。为了简单起见,有些话要说。只要您的所有表单输入都设置了name属性,这应该只适用于jimdandy。
$('form.myform').submit(函数(){var$this=$(this),viewArr=$this.serializeArray(),视图={};for(viewArr中的var i){view[viewArr[i].name]=viewArr[i].value;}//在这里使用视图对象(例如JSON.stringify?)});
这是对Tobias Cohen函数的改进,该函数在多维数组中运行良好:
http://jsfiddle.net/BNnwF/2/
然而,这不是一个jQuery插件,但如果您想这样使用它,只需几秒钟就可以将它变成一个:只需替换函数声明包装器:
function serializeFormObject(form)
{
...
}
具有:
$.fn.serializeFormObject = function()
{
var form = this;
...
};
我想这与梅斯克的解决方案相似,因为它做了相同的事情,但我认为这有点干净和简单。我还将macek的测试用例输入添加到小提琴中,并添加了一些额外的输入。到目前为止,这对我来说很好。
function serializeFormObject(form)
{
function trim(str)
{
return str.replace(/^\s+|\s+$/g,"");
}
var o = {};
var a = $(form).serializeArray();
$.each(a, function() {
var nameParts = this.name.split('[');
if (nameParts.length == 1) {
// New value is not an array - so we simply add the new
// value to the result object
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 || '';
}
}
else {
// New value is an array - we need to merge it into the
// existing result object
$.each(nameParts, function (index) {
nameParts[index] = this.replace(/\]$/, '');
});
// This $.each merges the new value in, part by part
var arrItem = this;
var temp = o;
$.each(nameParts, function (index) {
var next;
var nextNamePart;
if (index >= nameParts.length - 1)
next = arrItem.value || '';
else {
nextNamePart = nameParts[index + 1];
if (trim(this) != '' && temp[this] !== undefined)
next = temp[this];
else {
if (trim(nextNamePart) == '')
next = [];
else
next = {};
}
}
if (trim(this) == '') {
temp.push(next);
} else
temp[this] = next;
temp = next;
});
}
});
return o;
}