Php 如何列出成功上载的文件的名称?
我有一个包含文件输入和上传按钮的表单Php 如何列出成功上载的文件的名称?,php,javascript,jquery,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&g
<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>
PHP回调应该包含当前上载的文件名,甚至包含所有上载文件的数组 然后,在
stopmageupload
函数中,读取第二个参数并相应地操作#list
(更新/添加html):
函数stopImageUpload(成功,文件名){
var结果;
如果(成功==1)
结果=“文件“+filename+”已成功上载! ”;
其他的
结果=“上传文件“+filename+”时出错! ”;
$('#list')。追加(结果);
返回true;
}
我包含了将文件上传到服务器的php代码。你是说在这段代码中:window.top.window.stopImageUpload()代码>我需要回显<代码>$\u文件[“文件图像”][“名称”]
?是的,文件上传时会执行JS,不是吗?是的,调用了JS函数,所以如果我回显`$\u文件[“文件图像”][“名称”]`?在回调函数中,我只需要知道在javascript函数中如何将所有文件名添加到span标记“list”中?现在,这很简单。我在回答中添加了一个简单的例子,只是实现了您需要的。
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;
}
<script language="javascript" type="text/javascript">window.top.window.stopImageUpload(<?php echo $result; ?>);</script>
move_uploaded_file($_FILES["fileImage"]["tmp_name"],"ImageFiles/" . $_FILES["fileImage"]["name"]);
$result = 1;
<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;
}