Issue
I’ve tried answers from other posts, but they are not working for me.
Folders are: (py code is in ‘art’, flask is run from there)
art
static
css
html
img
js
lib
Import in html is:
<script src="{{url_for('static', filename='js/lib/jquery-3.4.1.slim.min.js')}}"> </script>
<script src="{{url_for('static', filename='js/Main.js')}}"> </script>
Error from Flask run:
127.0.0.1 - - [25/Aug/2021 10:46:17] "GET /art HTTP/1.1" 200 -
127.0.0.1 - - [25/Aug/2021 10:46:17] "GET /%7B%7Burl_for('static',%20filename='/js/bin/jquery-3.4.1.slim.min.js')%7D%7D HTTP/1.1" 404 -
127.0.0.1 - - [25/Aug/2021 10:46:17] "GET /%7B%7Burl_for('static',%20filename='/js/Main.js')%7D%7D HTTP/1.1" 404 -
From the Safari Web Inspector:
Thanks.
EDIT: .py code:
@app.route("/art")
def index():
return """
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<meta http-equiv="Expires" content="-1">
<title>CD Test</title>
<script src="{{url_for('static', filename='/js/lib/jquery-3.4.1.slim.min.js')}}"> </script>
<script src="{{url_for('static', filename='/js/Main.js')}}"> </script>
</head>
<body>
<h2>CD Test.</h2>
<hr>
<table>
<tr>
<td><img id="Pic" src="url/xxx.jpg" onclick="openFullscreen()"></td>
</tr>
</table>
</body>
</html>
"""
if __name__ == "__main__":
app.run(host="127.0.0.1", port=8080, debug=True)
Solution
This can not work. You can not return template tag (the pieces of code in {{ }}) directly. You have to render the template first. I suggest you go there for a quick tutorial/example on how to render a templates.
Answered By – Niko B
This Answer collected from stackoverflow, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0