我可以使用jquery ajax和php上传文件，需要一些说明和修改

我已经读过了。

https://developer.mozilla.org/en/docs/Web/API/FormData

仍然不明白为什么formdata（frm）在代码中为空。console.log(frm);返回“ FormData {}”为空。为什么？

file_form.php =>

<title>Upload File</title>
<div id="targetLayer">No Image</div>
<img id="my_image1" src="" />
<img id="my_image2" src="" />

<form id="uploadForm1" enctype="multipart/form-data">
    <input name="image1" type="file" />
    <input type="submit" value="Submit" />
</form>

<form id="uploadForm2" enctype="multipart/form-data">
    <input name="image2" type="file" />
    <input type="submit" value="Submit" />
</form>

<script type="text/javascript" src="jquery-1.11.3.js"></script>

<script type="text/javascript">

    $(document).ready(function (e) {
        $("#uploadForm1").on('submit', (function (e) {
            frm = new FormData($(this)[0]);
            console.log(frm);
            e.preventDefault();
            $.ajax({
                url: "upload_script1.php",
                type: "POST",
                data: frm,
                contentType: false,
                cache: false,
                processData: false,
                success: function (response) {
                    var JsonObject = JSON.parse(response);
                    $("#targetLayer").html(response);

                    var fl1 = "uploads/" + JsonObject.image1;
                    $("#my_image1").attr("src", fl1).height(100).width(100);
                }
            });
        }));

        $("#uploadForm2").on('submit', (function (e) {
            frm = new FormData($(this)[0]);
            e.preventDefault();
            $.ajax({
                url: "upload_script2.php",
                type: "POST",
                data: frm,
                contentType: false,
                cache: false,
                processData: false,
                success: function (response) {
                    var JsonObject = JSON.parse(response);
                    $("#targetLayer").html(response);

                    var fl2 = "uploads/" + JsonObject.image2;
                    $("#my_image2").attr("src", fl2).height(100).width(100);
                }
            });
        }));


    });
</script>

upload_script1.php =>

<?php
    $uploaddir = '/var/www/html/file-upload/uploads/';
    $uploadfile = $uploaddir . basename($_FILES['image1']['name']);
    move_uploaded_file($_FILES['image1']['tmp_name'], $uploadfile);

    $image1 = $_FILES['image1']['name'];
    $images["image1"] = $image1;
    echo json_encode($images);
?>

upload_script2.php =>

<?php
    $uploaddir = '/var/www/html/file-upload/uploads/';
    $uploadfile = $uploaddir . basename($_FILES['image2']['name']);
    move_uploaded_file($_FILES['image2']['tmp_name'], $uploadfile);
    $image2 = $_FILES['image2']['name'];
    $images["image2"] = $image2;
    echo json_encode($images);
?>

1. 任何人都可以将两个$.ajax()代码重写为一个$.ajax()代码，从而使代码更短吗？必要时重新编写html和php代码。

任何帮助将不胜感激。提前致谢。