form表单上传图片-提交给php方法,成功后想返回前台一些值,但前台始终无法获取到参数值。急!!!

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Release: 2016-06-02 11:31:12
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php图片表单upload

form表单部分代码:以此来实现form提交页面不跳转。

<code> <form action="user.php" method="post" enctype="multipart/form-data" target="id_iframe">                <input name="upfile" type="file" id="upfile">                <input name="act" type="hidden" value="uploadcardno">                <input type="submit" name="submit" id="upload" value="上传文件">            </form>            <iframe id="id_iframe" name="id_iframe" style="display:none;"></iframe></code>
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php方法:图片上传成功或失败,想返回图片的路径和报错信息。传值方式如下
$cardnoimg = $destination_folder; -- 图片路径
$smarty->assign('cardnoimg', $cardnoimg); -- 传入dwt模版页
$smarty->assign('mag', $mag); --错误信息传入dwt模版页
现在有两个问题
1:我想用JS来获取图片路径,并赋值给一个隐藏的文本框。那么这个方法应该在什么地方去写呢?在form的onsubmit方法里写不合适吧?
2:就是我php应该怎么写才能让dwt模版中读取到这些值呢?
本人菜鸟,请大神详细说明。

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