html5 拖拽上传，通过e.dataTransfer.files，拿到的file一直为null

Question

 <script type="text/javascript" charset="utf-8">
        //文件上传
        jQuery(function ($) {
            var uploadFile = function (file) {
                var element = $("<div />");
                element.text(file.fileName);

                var bar = $("<div />");
                element.append(bar);
                $("#progress").append(element);
                var onProgress = function (e) {
                    var per = Math.round((e.position / e.total) * 100);
                    bar.progressbar({ value: per });
                };
                var onSuccess = function () {
                    element.text("Complete");
                    element.delay(1000).fade();
                };

                $.upload("/Images/uploads", file, { upload: { progress: onProgress }, success: onSuccess });
            };

            // 创建拖拽目标区域
            window.addEventListener("load", function () {
                var element = document.getElementById("drop");

                element.addEventListener("dragenter", function (e) {
                    e.stopPropagation();
                    e.preventDefault();
                });

                element.addEventListener("dragover", function (e) {
                    e.dataTransfer.dropEffect = "copy";
                    e.stopPropagation();
                    e.preventDefault();
                });

                element.addEventListener("drop", function (e) {
                       var file = e.dataTransfer.files;   
                    for (var i = 0; i < file.length; i++)
                        alert("Drop " + file[i].name.toString());
                    uploadFile(file[i]);
                    return false;
                });
                document.body.addEventListener("dragover", function (e) {
                    e.stopPropagation();
                    e.preventDefault();
                    return false;
                });
            });
                   });

    </script>

Answer 1

应该是这样吧 var file = e.dataTransfer.files[0];