/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 vote134 views1 answer
Module level import not at top of file
Python
0 vote103 views1 answer
Inicair venv python
Python
8 votes67 views1 answer
Python string indexof
Python
2 votes104 views2 answers
Coding the fibonacci sequence
Python
67 votes109 views4 answers
List slicing reverse python
Python
6 votes131 views2 answers
Lofi hip hop radio online
Python
58 votes92 views6 answers
Matplotlib plot
Python
1 vote116 views1 answer
Convert number to time python
Python
0 vote89 views1 answer
Python pandas first and last element of column
Python
15 votes95 views6 answers
Google translation
Python