我是PHP初学者，我不知道如何从PHP视频文件中获得缩略图。我没有找到具体的解决办法。我尝试了ffmpeg库。但现在我想没有ffmpeg缩略图。所以请帮帮我。这是我的编码器代码。请审查。
<?php
$extension = pathinfo($_FILES["url"]["name"], PATHINFO_EXTENSION);/* Get file extension */
$filename = time() . '_' . mt_rand() . '.' . $extension;
move_uploaded_file($_FILES['url']['tmp_name'], $destinationPath . $filename);
$params['filename'] =$destinationPath.$filename;
$params['thumbImageName'] =$destinationPath.$thumbImageName;
$this->load->library('resize',$params);
$this->resize->resizeImage(WEIGHT, HEIGHT);
$this->resize->saveImage($destinationPath.$thumbImageName, 100);
?>
其实是有办法的。
上传方法:用户选择要上传的视频。选择之后，在弹出的div中显示一个加载消息，或者只使用一个警告框。然后，你要做的就是将视频文件加载到视频标签中。最好显示视频标签:无。待视频标签处于"就绪"状态后，跳到视频的50%。然后，获取视频标签的屏幕截图。你可以获取base64格式的图像，然后通过Ajax将其与视频一起发送到PHP。
包含base64脚本的输入标签。
<html>
<form action="" method="post" enctype="multipart/form-data" id="uploadvidform" accept="video/*">
<input type="file" name="file" id="file" />
<input type="text" name="screenshotbase64" id="screenshotbase64" value="" style="display: none;" />
</form>
<input type="button" value="Upload" id="uploadvidbut" />
和不可见的video元素。
<video width="400" id="videoelem" style="display: none;" controls>
<source src="" id="video_src">
Your browser does not support HTML5 video.
</video>
和脚本。首先，在文件被选中时执行一个操作。确保在head元素中链接了Google Ajax文件。
<script>
var scalefac = 0.25;
// Scale of image;
var screenshots = [];
// An array for multiple screenshots;
// This function will create an image. It's not used now, it's used in the below action (when you change the file).
function capture(video, scalefac) {
if(scaleFactor == null){
scaleFactor = 1;
}
var w = video.videoWidth * scalefac;
var h = video.videoHeight * scalefac;
var canvas = document.createElement('canvas');
canvas.width = w;
canvas.height = h;
var ctx = canvas.getContext('2d');
ctx.drawImage(video, 0, 0, w, h);
return canvas;
}
$(document).ready(function(){
$(document).on("change", "#file", function(){
alert("Please wait while we verify your video. This will only take a couple of seconds.");
// The next 3 lines will load the video
var lasource = $('#video_src');
lasource[0].src = URL.createObjectURL($('#file').prop("files")[0]);
lasource.parent()[0].load();
var video = document.getElementById("videoelem");
setTimeout(function(){
// Video needs to load then we check the state.
if (video.readyState == "4"){
var videoduration = $("#videoelem").get(0).duration;
var timetogoto = videodurationinseconds / 2;
$("#videoelem").get(0).currentTime = timetogoto;
setTimeout(function(){
// Video needs to load again
var video = document.getElementById("videoelem");
// function the screen grab.
var canvas = capture(video, scalefac);
screenshots.unshift(canvas);
for(var i=0; i<4; i++){
$("#screenshotbase64").val(screenshots[i].toDataURL());
}
}, 500);
}, 3000);
});
// Now that the form is filled, you can send your data to your PHP file.
$(document).on('click', '#uploadvidbut', function(){
var form = new FormData($("#uploadvidform")[0]);
$.ajax({
url: '/uploadvideodocument.php', // PHP file
type: 'POST',
data: form,
cache: false,
contentType: false,
processData: false,
success: function (result){
if (result == 1){
alert("The video has been uploaded.");
}
}
}).fail(function(){
alert("Oh no, the video wasn't uploaded.");
});
});
});
</script>
现在是PHP文件。我只打算包括base64转换为图像，我希望你知道如何做剩下的。
<?php
$data = $_POST['screenshotbase64'];
list($type, $data) = explode(';', $data);
list(, $data) = explode(',', $data);
$data = base64_decode($data);
// The following 2 lines will create the time in microseconds which you can use as the name. Microseconds ensures an almost impossibility of two people uploading at the same time.
$mt = explode(' ', microtime());
$millies = ((int)$mt[1]) * 1000 + ((int)round($mt[0] * 1000));
$screenshotfilename = time(). $millies . '.png';
// In the next line, replace YOUR DIRECTORY with your home path and then include the folder of where you want to save the screenshot.
file_put_contents('YOUR DIRECTORY AND FOLDER' . $screenshotfilename, $data);
// Now, the screen shot has been saved to the server and the name of the file is $screenshotfilename.
?>
注意:
有些浏览器可能不接受video元素。如今，这种情况几乎不会发生。但是一定要记住。