Javascript 如何在通过onChange（）检查图像文件后使用jquery提交表单？

这是我的表格-

<form id="imgUpIcn" class="imgUpIcn" enctype="multipart/form-data" method="post" action="myfile.php">
<div id="upload-file-container">
<input type="file" name="uploadedImagefile" id="myImageFile" onchange="photoUpInit(event,this,'loadingImg','cropPhLayer')" />
</div>   
</form>  

现在，当执行表单操作时，我的文件将上载到服务器上，但我希望从php文件获得回调，而不是重定向到php页面

这是我的php文件-

<?php
if (isset($_FILES['uploadedImagefile']['tmp_name']))
{


$target_path = "../uploads/";

$target_path = $target_path . "tmp"; 

if(move_uploaded_file($_FILES['uploadedImagefile']['tmp_name'], $target_path)) {
    echo "The file ".  basename( $_FILES['uploadedImagefile']['name']). 
    " has been uploaded";
} else{
    echo "There was an error uploading the file, please try again!";
}
}

---

考虑到是php文件上传了您的文件，您应该将php头重定向回表单页面，并显示成功或失败的消息。大概是这样的：

<?php
if (isset($_FILES['uploadedImagefile']['tmp_name']))
{


$target_path = "../uploads/";

$target_path = $target_path . "tmp"; 

if(move_uploaded_file($_FILES['uploadedImagefile']['tmp_name'], $target_path)) {
    $message = "The file ".  basename( $_FILES['uploadedImagefile']['name']). 
    " has been uploaded";
} else{
    $message = "There was an error uploading the file, please try again!";
}

header('Location: back/to/your/form/file?message=' . $message);
}

---

我没有类似问题的提交按钮。我想在Ajax成功函数中获得结果。然后需要使用另一种方法，通过Ajax将图像发送到php脚本，称为ajaxupload。谷歌它，因为我们不允许提供图书馆的答案。
$('#imgUpIcn').on('submit', (function(e) {
    e.preventDefault();
    $.ajax({
        type : "POST",
        url : $(this).attr('action'),
        data : $(this).serialize(),
        success : function(data) {
            $("#myModal").dialog("close");
        },
        error : function(data) {
            console.log("error");
            console.log(data);
        }
    });
}));

<?php
if (isset($_FILES['uploadedImagefile']['tmp_name']))
{


$target_path = "../uploads/";

$target_path = $target_path . "tmp"; 

if(move_uploaded_file($_FILES['uploadedImagefile']['tmp_name'], $target_path)) {
    $message = "The file ".  basename( $_FILES['uploadedImagefile']['name']). 
    " has been uploaded";
} else{
    $message = "There was an error uploading the file, please try again!";
}

header('Location: back/to/your/form/file?message=' . $message);
}