如何列出成功上传的文件名?

时间:2012-04-16 15:56:41

标签: php javascript jquery

我有一个包含文件输入和上传按钮的表单

<form action='imageupload.php' method='post' enctype='multipart/form-data' target='upload_target' onsubmit='stopImageUpload(this);' class='imageuploadform' >
    <p class='imagef1_upload_form' align='center'><label>Image File: <input name='fileImage' type='file' class='fileImage' /></label>
        <input type='submit' name='submitImageBtn' class='sbtnimage' value='Upload' /></label>
    </p>
    <p><span class='list></span></p>
</form>

现在我想要的是,如果图像已经完成上传,则在此表单的底部(在span标签'list'中)我想要已经上传的文件的名称。因此,当文件成功上传时,它应显示上传文件的名称。如果第二个文件成功上载,则它会在第一个文件的名称下方显示上载的第二个文件的名称。 这可能在Javascript中吗?

我上传的文件全部用php和javascript整理出来。以下是上传完成后的javascript功能:

function stopImageUpload(success) {
    var result = '';
    if (success == 1) {
        result = '<span class="msg">The file was uploaded successfully!</span><br/><br/>';
    }
    else {
        result = '<span class="emsg">There was an error during file upload!</span><br/><br/>';
    };
    return true;
}​

此功能位于fomr的同一页面,但与上传脚本不在同一页面中,但在上传脚本中使用以下javascript代码调用此功能:

<script language="javascript" type="text/javascript">window.top.window.stopImageUpload(<?php echo $result; ?>);</script> 

下面是上传php代码的一部分,它将文件上传到服务器:

move_uploaded_file($_FILES["fileImage"]["tmp_name"],"ImageFiles/" . $_FILES["fileImage"]["name"]);
$result = 1;

1 个答案:

答案 0 :(得分:0)

您的PHP回调应该包含当前上传文件的名称,甚至包含所有上传文件的数组。

然后,在stopImageUpload函数中,读取第二个参数并按照{更新/添加html}操作#list

<ul id="list"></ul>
function stopImageUpload(success, filename) {
    var result;
    if (success == 1)
        result = '<li class="msg">The file '+filename+' was uploaded successfully!</li>';
    else
        result = '<li class="emsg">There was an error during upload of file '+filename+'!</li>';
    $('#list').append(result);
    return true;
}​