Creating dynamic flask route-CodePudding

I'm attempting to create a dynamic flask route. I want to create a parallel page for my route that internal users can see when they add 'vo/' before the URL. So essentially there are two different routes /vo/feedbackform and /feedbackform. I know I could just create two routes in the app, but I'm hoping I can optimize.

@app.route('/<x>feedbackform', methods=['GET', 'POST'])
def feedback_portal(x):
    if x == 'vo/':
        return render_template('feedback_vo.html', title='Inquiry Submission')
    elif x == '':
        return render_template('feedback.html', title='Inquiry Submission')

So far, this only works when the URL using the vo/ in the x part of the URL by my elif isn't working, and I get an error.

CodePudding user response：

i dont think this is supported by Flask.

For your case that you have only 2 paths to support, you can add two decorators/routes to same function and then inspect the request.path to render the appropriate template:

from flask import request

@app.route('/vo/feedbackform', methods=['GET', 'POST'])
@app.route('/feedbackform', methods=['GET', 'POST'])
def feedback_portal():
    if request.path[0:4] == '/vo/':
        return render_template('feedback_vo.html', title='Inquiry Submission')
    else:
        return render_template('feedback.html', title='Inquiry Submission')

if you had multiple options to support instead of just /vo/, you could look into declaring the dynamic part as path:

@app.route('/<path:x>feedbackform', methods=['GET', 'POST'])

And add one more route to handle the static /feedbackform.

CodePudding user response：

Building off of @ilias-sp's answer, I was able to use this code:

@app.route('/feedbackform/', methods=['GET', 'POST'], defaults={'x': None})
@app.route('/<x>/feedbackform/', methods=['GET', 'POST'])
def feedback_portal(x):
    if x=='vo':
        return render_template('feedback_vo.html', title='Inquiry Submission')
    elif x==None:
        return render_template('feedback.html', title='Inquiry Submission')