我怎样才能将我的JS对象转换为FormData?

我这样做的原因是,我有一个用~100个表单字段值构造的对象。

var item = {
   description: 'Some Item',
   price : '0.00',
   srate : '0.00',
   color : 'red',
   ...
   ...
}

现在我被要求将上传文件功能添加到我的表单,当然,通过JSON是不可能的,所以我计划移动到FormData。那么有什么方法可以将我的JS对象转换为FormData呢?


当前回答

我参考了古德拉丹的回答。我用Typescript格式编辑了一下。

class UtilityService {
    private appendFormData(formData, data, rootName) {

        let root = rootName || '';
        if (data instanceof File) {
            formData.append(root, data);
        } else if (Array.isArray(data)) {
            for (var i = 0; i < data.length; i++) {
                this.appendFormData(formData, data[i], root + '[' + i + ']');
            }
        } else if (typeof data === 'object' && data) {
            for (var key in data) {
                if (data.hasOwnProperty(key)) {
                    if (root === '') {
                        this.appendFormData(formData, data[key], key);
                    } else {
                        this.appendFormData(formData, data[key], root + '.' + key);
                    }
                }
            }
        } else {
            if (data !== null && typeof data !== 'undefined') {
                formData.append(root, data);
            }
        }
    }

    getFormDataFromObj(data) {
        var formData = new FormData();

        this.appendFormData(formData, data, '');

        return formData;
    }
}

export let UtilityMan = new UtilityService();

其他回答

你可以简单地使用:

formData.append('item', JSON.stringify(item));

我有一个场景,在构造表单数据时,嵌套的JSON必须以线性方式序列化,因为这是服务器期望值的方式。所以,我写了一个小的递归函数来翻译JSON,就像这样:

{
   "orderPrice":"11",
   "cardNumber":"************1234",
   "id":"8796191359018",
   "accountHolderName":"Raj Pawan",
   "expiryMonth":"02",
   "expiryYear":"2019",
   "issueNumber":null,
   "billingAddress":{
      "city":"Wonderland",
      "code":"8796682911767",
      "firstname":"Raj Pawan",
      "lastname":"Gumdal",
      "line1":"Addr Line 1",
      "line2":null,
      "state":"US-AS",
      "region":{
         "isocode":"US-AS"
      },
      "zip":"76767-6776"
   }
}

变成这样:

{
   "orderPrice":"11",
   "cardNumber":"************1234",
   "id":"8796191359018",
   "accountHolderName":"Raj Pawan",
   "expiryMonth":"02",
   "expiryYear":"2019",
   "issueNumber":null,
   "billingAddress.city":"Wonderland",
   "billingAddress.code":"8796682911767",
   "billingAddress.firstname":"Raj Pawan",
   "billingAddress.lastname":"Gumdal",
   "billingAddress.line1":"Addr Line 1",
   "billingAddress.line2":null,
   "billingAddress.state":"US-AS",
   "billingAddress.region.isocode":"US-AS",
   "billingAddress.zip":"76767-6776"
}

服务器将接受这种转换格式的表单数据。

函数如下:

function jsonToFormData (inJSON, inTestJSON, inFormData, parentKey) {
    // http://stackoverflow.com/a/22783314/260665
    // Raj: Converts any nested JSON to formData.
    var form_data = inFormData || new FormData();
    var testJSON = inTestJSON || {};
    for ( var key in inJSON ) {
        // 1. If it is a recursion, then key has to be constructed like "parent.child" where parent JSON contains a child JSON
        // 2. Perform append data only if the value for key is not a JSON, recurse otherwise!
        var constructedKey = key;
        if (parentKey) {
            constructedKey = parentKey + "." + key;
        }

        var value = inJSON[key];
        if (value && value.constructor === {}.constructor) {
            // This is a JSON, we now need to recurse!
            jsonToFormData (value, testJSON, form_data, constructedKey);
        } else {
            form_data.append(constructedKey, inJSON[key]);
            testJSON[constructedKey] = inJSON[key];
        }
    }
    return form_data;
}

调用:

        var testJSON = {};
        var form_data = jsonToFormData (jsonForPost, testJSON);

我使用testJSON只是为了查看转换后的结果,因为我无法提取form_data的内容。AJAX post call:

        $.ajax({
            type: "POST",
            url: somePostURL,
            data: form_data,
            processData : false,
            contentType : false,
            success: function (data) {
            },
            error: function (e) {
            }
        });

递归地

const toFormData = (f => f(f))(h => f => f(x => h(h)(f)(x)))(f => fd => pk => d => { if (d instanceof Object) { Object.keys(d).forEach(k => { const v = d[k] if (pk) k = `${pk}[${k}]` if (v instanceof Object && !(v instanceof Date) && !(v instanceof File)) { return f(fd)(k)(v) } else { fd.append(k, v) } }) } return fd })(new FormData())() let data = { name: 'John', age: 30, colors: ['red', 'green', 'blue'], children: [ { name: 'Max', age: 3 }, { name: 'Madonna', age: 10 } ] } console.log('data', data) document.getElementById("data").insertAdjacentHTML('beforeend', JSON.stringify(data)) let formData = toFormData(data) for (let key of formData.keys()) { console.log(key, formData.getAll(key).join(',')) document.getElementById("item").insertAdjacentHTML('beforeend', `<li>${key} = ${formData.getAll(key).join(',')}</li>`) } <p id="data"></p> <ul id="item"></ul>

使用jquery,你可以通过$.param(obj)简单地做到这一点。

例子: Const obj = { 描述:'Some Item', 价格:0.00, srate:“0.00”, 颜色:红色的 } Const form_obj = $.param(obj); . ajax({美元 url:“example.com”, 方法:“文章”, 数据:form_obj }) < script src = " https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js " > < /脚本>

如果您有一个对象,您可以轻松地创建一个FormData对象,并将该对象的名称和值附加到FormData。

你还没有发布任何代码,所以这是一个通用的例子;

var form_data = new FormData();

for ( var key in item ) {
    form_data.append(key, item[key]);
}

$.ajax({
    url         : 'http://example.com/upload.php',
    data        : form_data,
    processData : false,
    contentType : false,
    type: 'POST'
}).done(function(data){
    // do stuff
});

在MDN的文档中有更多的例子