/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

1 vote316 views1 answer
TypeError: ‘int’ object is not subscriptable Solution
Python
0 vote361 views1 answer
Type object 'SparkContext' has no attribute '_jsc'
Python
1 vote326 views1 answer
Extract url from page python
Python
2 votes352 views1 answer
Add header to table in pandas
Python
2 votes317 views1 answer
Creating a 50 day and 100 day moving average python
Python
1 vote323 views1 answer
Python clamp
Python
13 votes342 views6 answers
Range parameters python
Python
2 votes346 views2 answers
Python Decompress gzip File
Python
7 votes328 views3 answers
Pandas empty dataframe with column names
Python
0 vote331 views1 answer
Python map function using lambda function as one of the parameters
Python