I have a flask template that has a video like:
<video id="firstVideo" class="clip-thumbnail" src=""></video>
and upon an update from a user click, the request's videos content updates the video like:
$('#firstVideo').attr("src", response['videos'][0]);
but I want this to be dynamically updating on the page with something like:
$('#firstVideo').attr("src", {{ url_for('static', filename='videos/' + response['videos'][0] ) }} )
Is this the right way to handle this? Will flask know to re-render the page after the update or do I need to re-render the page?
除非您在服务器上构建动态模板,否则不能使用flask模板标签,如果要从浏览器向视频URL发出ajax请求,则只需设置请求以返回完整的视频URL,然后然后完全按照您的要求更新视频src:
$('#firstVideo').attr("src", response['videos'][0]);
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