Question

当我 console.log(formData); 为空时，

以下是我的代码。看不出它失败的原因。

$("#add_meal_item").on('click', '#save_meal_item', function () {
                var meal_name = $("#add_meal_item_form").find('#meal_name').val();
                var meal_type = $("#add_meal_item_form").find('#meal_type').val();
                var logoImg = $("#add_meal_item_form").find('#meal_image').get(0).files[0];
                var formData = new FormData();
                formData.append('logo', logoImg);
                formData.append('meal_type', meal_type);
                formData.append('meal_name', meal_name);
                console.log(formData);
                var ajax_url = baseurl + '/save_meal_item';
                $.ajax({
                    url: ajax_url,
                    type: 'POST',
                    data: formData,
                    success: function (response) {

                    }
                });

            });

Answer 1

您需要先获取元素值，然后将它们附加到formData，如下所示：

var meal_name = $("#add_meal_item_form").find('#meal_name').val();
formData.append('meal_name ', meal_name );

要显示所有formData键和值，您必须使用FormData.entries方法：

for(var pair of formData.entries()) {
     console.log(pair[0]+ ', '+ pair[1]); 
}

Answer 2

FormData.append()不会为formData创建可枚举属性，您需要使用FormData.entries中的迭代器来迭代这些值。

<强>演示

var formData = new FormData();
formData.append('logo', "11");
formData.append('meal_type', "22");
formData.append('meal_name', "33");

console.log(Object.keys(formData)); //[]
console.log(JSON.stringify(formData)); //{}

//use iterator
var formEntries = formData.entries();
//console.log(formEntries.next().value); 
//console.log(formEntries.next().value); 
//console.log(formEntries.next().value); 

//or simply use Array.from
var formEntries = Array.from(formData.entries());
console.log("formEntries " , formEntries);

尝试在其中附加值时，Javascript formdata为空

2 个答案: