/Python/Flask/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 vote204 views2 answers
Flask abort
Python Flask
1 vote207 views1 answer
Flask return error response
Python Flask
1 vote194 views2 answers
Flask socketio send
Python Flask
2 votes219 views1 answer
Check if user log in flask
Python Flask
0 vote189 views1 answer
Mail.send_message flask not working, SSL ==> 465
Python Flask
1 vote198 views1 answer
Flask define template folder
Python Flask
0 vote212 views1 answer
How to return an html file in flask
Python Flask
8 votes220 views2 answers
Flask.pocoo.org quickstart
Python Flask
0 vote208 views3 answers
Flask models
Python Flask
0 vote213 views1 answer
Gevent with flask
Python Flask