Question

不发送给任何东西。当我试图在表单数据上显示结果时，它看起来是空的。

我的表格：

    <form id="uploadform1236-1" method="POST" enctype="multipart/form-data" onsubmit="uploadprove('uploadform1236-1', event, 'post.php?action=uploadprove');">
<input name="picture" id="file-upload1236-1" class="" type="file">
<input name="testx" id="xxx1" class="" value="test1236" type="text">
<button id="uploadbutton1236-1" class="btn btn-primary btn-sm" type="submit">ttt</button>

</form>

使用Javascript：

function uploadprove(id, e, url) {
        e.preventDefault();
        var uploadform = $('#' + id)[0];
        var formData = new FormData(document.getElementById(id)); //also tried to put uploadform variable here
        alert(uploadform.toSource()); // results are ({0:{},1:{},2:{}}) It sees that 3 elements in form
        alert(formData.toSource()); // results are ({})
        $.ajax({
            url: url,
            type: "POST",
            data: formData,
            ContentType: false,
            ProcessData: false,
            success: function (data) {


                alert(data);


            }
        })
    }

Answer 1

您的字段ContentType和ProcessData有误，是contentType和processData

$.ajax({
    url: url,
    type: "POST",
    data: formData,
    contentType: false,
    processData: false,
    success: function (data) {
        alert(data);
    }
})

ajax上传文件formdata看起来是空的

1 个答案: