/Python/Flask conditional according to urrl

Flask conditional according to urrl

<!DOCTYPE html>
<html>
<head>
    <title>Template</title>
</head>
<body>
    <h3>You are here: {{ request.path }}</h3>
    {% if request.path == url_for('show_funding') %}
        <script src="{{ url_for ('static', filename='js/funding.js')}}"></script>
    {% endif %}
</body>
</html>

Source: stackoverflow.com

Flask conditional according to urrl

from flask import Flask
from flask import request
from flask import render_template

app = Flask(__name__)

@app.route('/')
def show_index():
    return render_template('funding.html')

@app.route('/funding')
def show_funding():
    return render_template('funding.html')

if __name__ == '__main__':
    app.run(debug = True)

Source: stackoverflow.com

Similar codes

0 vote216 views1 answer
How to close turle loop
Python
41 votes250 views6 answers
How to remove rows with nan in pandas
Python
0 vote217 views4 answers
Mergesort
Python
-1 vote210 views1 answer
"scrapy shell" pass cookies to fetch
Python
79 votes232 views4 answers
Write a file python
Python
0 vote224 views1 answer
Socket python how to check if server alive
Python
7 votes225 views1 answer
Python title case
Python
0 vote219 views1 answer
Online python formatter and compiler
Python
0 vote224 views1 answer
Tryexept in python
Python
0 vote201 views1 answer
Change xlabel python
Python