Php 如何在文本输入中显示变量
在Php 如何在文本输入中显示变量,php,javascript,jquery,Php,Javascript,Jquery,在qanadable.php中,我有一个包含文件输入和iframe的表单: var $fileVideo = $("<form action='videoupload.php' method='post' enctype='multipart/form-data' target='upload_target_video' onsubmit='return videoClickHandler(this);' class='videouploadform' >" + "Video Fi
qanadable.php
中,我有一个包含文件输入和iframe的表单:
var $fileVideo = $("<form action='videoupload.php' method='post' enctype='multipart/form-data' target='upload_target_video' onsubmit='return videoClickHandler(this);' class='videouploadform' >" +
"Video File: <input name='fileVideo' type='file' class='fileVideo' /></label>" +
"<input type='submit' name='submitVideoBtn' class='sbtnvideo' value='Upload' /></label>" +
"<p class='listVideo' align='left'></p>" +
"<iframe class='upload_target_video' name='upload_target_video' src='/' style='width:0px;height:0px;border:0px;solid;#fff;'></iframe></form>");
我可能误解了你的问题,但如果你在输入字段中将
type='text'
更改为type='hidden'
,那么$id变量将位于隐藏输入中。对不起,我的问题太仓促了,我的意思是将其转换为文本输入,而不是隐藏输入,很抱歉,你不能将javascript和php混合使用,如果这是你要问的。对于javascript(客户端)和php(服务器端)之间的通信,必须使用AJAX-。另外,请注意,可以使用javascript库简化AJAX请求,如jQuery@intelis我一直试图效仿W3school的例子,但似乎无法让它发挥作用。你能提供样品吗?
function stopVideoUpload(success, videofilename){
var result = '';
videocounter++;
if (success == 1){
result = '<span class="videomsg'+videocounter+'">The file was uploaded successfully</span>';
$('.listVideo').eq(window.lastUploadVideoIndex).append('<div>' + htmlEncode(videofilename));
}
return true;
}
<script language="javascript" type="text/javascript">
window.top.stopVideoUpload(<?php echo $result; ?>,'<?php echo $id . $_FILES['fileVideo']['name'] ?>');
</script>