我需要在特定目录中上传视频，成功上传后，它会显示在页面上并带有缩略图预览：
代码：
<form action="" method="post" enctype="multipart/form-data">
<input type="hidden" name="MAX_FILE_SIZE" value="500000" />
<table width="400" cellpadding="3" >
<tr>
<td colspan="3"> </td>
</tr>
<tr>
<td width="10" rowspan="2"> </td>
<td width="120"><strong>Choose a file to upload:</strong></td>
<td width="242"><input type="file" name="uploaded_file" /></td>
</tr>
<tr>
<td> </td>
<td> </td>
</tr>
<tr>
<td> </td>
<td> </td>
<td><input type="submit" name="sendForm" value="Upload File" />
<br /></td>
</tr>
<tr>
<td colspan="3"> </td>
</tr>
</table>
</form>
<?php
// C:\wamp\www\scriptDir\uploads\;
$idir = "C:/xampp/htdocs/xampp/video"; // Path To Images Directory
$tdir = "C:/xampp/htdocs/xampp/video"; // Path …我需要根据选中的复选框获取值...使用HTML和jQuery或Javascript我尝试过以下代码:
jQuery代码:
<script type="text/javascript">
$(function () {
$('input:not(#submit)').click(function () {
t = $(this).attr('id');
text = $('.time' + t).text();
//alert(text);
});
$('#submit').click(function (e) {
e.preventDefault();
var trs = $('table tr');
var values = trs.first().find('td');
var values1 = $('table tr td :checkbox:checked').map(function () {
return $(this).closest('tr').find('td').text() + "==>"
+ values.eq($(this).parent().index()).text();
}).get();
alert(values1);
});
});
</script>
HTML标记:
<body>
<table border="1">
<tr>
<td>No</td>
<td >Name</td>
<td >City</td>
<td >Check</td>
</tr>
<tr>
<td>1</td>
<td>
select</td>
<td>
select</td>
<td>
<input type="checkbox" name="checkbox" class="r1" />select</td> …