Answers for "flask url_for"

4

default flask app

from flask import Flask
app = Flask(__name__)

@app.route('/')
def index():
    return 'Flask'
Posted by: Guest on June-13-2020
1

flask api with parameter

args = request.args
print (args) # For debugging
no1 = args['key1']
no2 = args['key2']
return jsonify(dict(data=[no1, no2])) # or whatever is required
Posted by: Guest on November-08-2020

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