【发布时间】:2015-04-19 13:30:28
【问题描述】:
我正在尝试使用 ajax -ffmpeg 制作视频上传系统。但我有一个问题。问题是我的 php 代码每次都说 Invailed file video Format。我在哪里做错了,我不明白。
有人可以帮我吗?
<?php
include_once 'includes.php';
include_once'includes/alphaID.php';
include_once 'includes/getExtension.php';
$valid_formats = array("mp4", "MP4");
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST")
{
$name = $_FILES['videop']['name'];
$size = $_FILES['videop']['size'];
if(strlen($name))
{
$ext = strtolower(pathinfo($video_numid, PATHINFO_EXTENSION));
$video_numid = alphaID(microtime(true) * 10000);
if(in_array($ext,$valid_formats))
{
if($size<(50024*50024))
{
$actual_video_name = time().$uid.".".$ext;
$tmp = $_FILES['videop']['tmp_name'];
$video_numid = $_FILES['file']['name'];
if(move_uploaded_file($tmp, $path.$actual_video_name))
{
shell_exec("ffmpeg -i $path.$actual_video_name.flv -f flv -s 650x390 $path.$actual_video_name.mp4");
shell_exec("ffmpeg -i $path.$actual_video_name.mp4 -vcodec png -ss 00:00:15 -s 650x390 -vframes 1 -an -f rawvideo $path.$actual_video_name.png");
$data=$Swamp->Video_Upload($uid,$actual_video_name);
$newdata=$Swamp->Get_Upload_Video($uid,$actual_video_name);
if($newdata)
{
echo '<div class="video_prew"><img src="'.$base_url.$path.$actual_video_name.'" width="100%" height="auto" class="preview_video" id="'.$newdata['id'].'"/></div>';
}
}
else
{
echo "Fail upload folder with read access.";
}
}
else
echo "Video file size max 10 MB";
} else
echo "Invalid file video format.";
} else
echo "Please select video..!";
exit;
}
?>
【问题讨论】:
-
为什么要依赖扩展来查看支持的格式?
-
是的,你是对的,我做错了,谢谢 :)