当我尝试从我刚刚创建的视频中捕获帧时出现错误。 它在 drawImage 之后:
Failed to execute 'drawImage' on 'CanvasRenderingContext2D': The provided value is not of type '(CSSImageValue or HTMLImageElement or SVGImageElement or HTMLVideoElement or HTMLCanvasElement or ImageBitmap or OffscreenCanvas)'
知道可能是什么原因吗?谢谢
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
var $theImg = $('<video id="video" preload="metadata" src="http://localhost/screens/page/VIDEO/vids/' + file.name + '"></video>');
$theImg.attr('data-disp', name);
$theImg.attr('data-name', file.name);
$theImg.attr('data-created', 1);
$theImg.appendTo('#selectable-videos');
$theImg.addClass('selectable-image');
$(e.target).parent().parent().find('.modal-body').children().append($theImg);
var $canvas = $('<canvas id="canvas" width="640" height="480"></canvas>');
$theImg.append($canvas);
var canvas = $('#canvas');
var video = $('#video');
if (video.length) {
var ctx = canvas[0].getContext('2d').drawImage(video, 0, 0, canvas.width, canvas.height);
//convert to desired file format
var dataURI = canvas.toDataURL('image/jpeg'); // can also use 'image/png'
最佳答案
您正在向 drawImage()
证明一个 jQuery 对象,而不是 native Element 对象。您可以使用 [0]
或 get(0)
从 jQuery 对象中检索第一个底层元素:
var ctx = canvas[0].getContext('2d').drawImage(video[0], 0, 0, canvas.width, canvas.height);
另请注意,当您在 Canvas 上调用 toDataURL()
时,您将在下一行遇到类似的问题。
var dataURI = canvas[0].toDataURL('image/jpeg');
请注意,可以整理逻辑,因为使用相似的变量名保存不同的对象类型会有点困惑。试试这个:
var $video = $('<video></video>', {
id: 'video',
preload: 'metadata',
src: 'http://localhost/screens/page/VIDEO/vids/' + file.name,
data - disp: name,
data - name: file.name,
data - created: 1,
class: 'selectable-image'
}).appendTo('#selectable-videos');
// is this line really needed? You just appended it above.
// also, use closest() instead of parent().parent().parent()...
$(e.target).parent().parent().find('.modal-body').children().append($video);
var $canvas = $('<canvas></canvas>', {
id: 'canvas',
width: 640,
height: 480
}).appendTo($video);
var canvas = $canvas[0];
var video = $video[0];
// removed 'if' check - you just created the element, it will exist
var ctx = canvas.getContext('2d').drawImage(video, 0, 0, canvas.width, canvas.height);
var dataURI = canvas.toDataURL('image/jpeg');
最后,您将 canvas
附加到 video
元素是非常奇怪的。我建议不要那样做。
关于javascript - 将图像从视频 JS 保存到 Canvas ,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54647582/