问如何通过jQuery步骤上传图片

EN

<input type="file" name="upload_doc" id="upload_doc" title="Search for a file to add"/>
<input id="uploadImage" type="button" value="Upload Image" name="upload"/>


<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.4/jquery.min.js"></script>
 <script type="text/javascript">
    jQuery.noConflict();
    jQuery('#uploadImage').on("click", function (e) {
        var uploadedFile = new FormData();
        uploadedFile.append('upload_doc', upload_doc.files[0]);
        jQuery.ajax({
          url: 'lab1.php',
          type: 'POST',
          processData: false, // important
          contentType: false, // important
          dataType : 'json',
          data: uploadedFile
        });
    });
</script>

<?php
    // check record array
    print_r($_FILES);
    move_uploaded_file($_FILES['upload_doc']['tmp_name'], $_FILES['upload_doc']['name']);
?>

<form action="" name="imageUploadForm" id="imageUploadForm" method="post" enctype="multipart/form-data">
<input type="file" style="background-color: #2184b3; color: #fff;" class="btn btn-default" name="upload_doc" id="upload_doc" title="Search for a file to add">
</form>
<script type="text/javascript">
    $(document).ready(function (e) {
        $('#imageUploadForm').on('submit',(function(e) {
            e.preventDefault();
            var formData = new FormData(this);

            $.ajax({
                type:'POST',
                url: $(this).attr('action'),
                data:formData,
                cache:false,
                contentType: false,
                processData: false,
                success:function(data){
                    console.log("success");
                    console.log(data);
                },
                error: function(data){
                    console.log("error");
                    console.log(data);
                }
            });
        }));

        $("#upload_doc").on("change", function() {
            $("#imageUploadForm").submit();
        });
    });
</script>

if(currentIndex == 0){
    event.preventDefault();
    var formData = new FormData($('form')[0]);
    $.ajax({
        async: false,
        url: myUrl,
        dataType: 'json',
        type: 'POST',
        cache: false,
        contentType: false,
        processData: false,
        data : formData,
        success: function(response) {
            console.log(response);
        }
    });
}