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
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);