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flask conditional according to urrl

Defeated Dogfish answered on November 26, 2020 Popularity 3/10 Helpfulness 1/10

answer flask conditional according to urrl

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flask conditional according to urrl

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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)

Popularity 3/10 Helpfulness 1/10 Language python

Source: stackoverflow.com

Tags: conditional flask python

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Contributed on Nov 26 2020

Defeated Dogfish

0 Answers Avg Quality 2/10

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flask conditional according to urrl

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<!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>

Popularity 3/10 Helpfulness 1/10 Language python

Source: stackoverflow.com

Tags: python flask conditional

Link to this answer
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Contributed on Nov 26 2020

Defeated Dogfish

0 Answers Avg Quality 2/10

flask conditional according to urrl

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