How to pass a URL parameter using python, Flask, and the command line

Question

I am having a lot of trouble passing a URL parameter using request.args.get.

My code is the following:

from flask import Flask, request
app= Flask(__name__)

@app.route('/post?id=post_id', methods=["GET"])
def show_post():
    post_id=1232
    return request.args.get('post_id')

if __name__=="__main__":
     app.run(host='0.0.0.0')

After saving, I always type python filename.py in the command line, see Running on http://0.0.0.0:5000/ (Press CTRL+C to quit) as the return on the command line, and then type in the url (http://ip_addres:5000/post?id=1232) in chrome to see if it will return 1232, but it won't do it! Please help.

bdjett · Accepted Answer · 2016-07-22 06:34:21Z

12

You should leave the query parameters out of the route as they aren't part of it.

@app.route('/post', methods=['GET'])
def show_post():
    post_id = request.args.get('id')
    return post_id

request.args is a dictionary containing all of the query parameters passed in the URL.

edited Jul 22, 2016 at 6:34

answered Jul 22, 2016 at 6:24

bdjett

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5 Comments

SAS Over a year ago

What would the URL be then, please? Thank you so much!

bdjett Over a year ago

The URL would still be http://<IP Address>:5000/post?id=1232. id is a query parameter that will appear in the request.args dictionary with the value 1232. Note I edited the above answer to correct the name of the parameter.

SAS Over a year ago

so basically, the request.args.get('id') automatically inserts the param inside the url?

bdjett Over a year ago

No, not at all. You enter the URL http://127.0.0.1:5000/post?id=1232. Flask will cut off the query string, which is ?id=1232 and match on the actual path, which is /post. That path gets routed to the show_post function. request.args is a dictionary containing all of the query parameters that were passed along with the URL, which in this case is the key id and the value 1232. request.args here will be a dictionary that looks like this: {"id": "1232"}.

Raja Simon Over a year ago

@SAS please see my answer also

Raja Simon · Accepted Answer · 2016-07-22 06:44:52Z

6

request.args is what you are looking for.

@app.route('/post', methods=["GET"])
def show_post()
    post_id = request.args.get('id')

request.args is get all the GET parameters and construct Multidict like this MultiDict([('id', '1232'),]) so using get you can get the id

edited Jul 22, 2016 at 6:44

answered Jul 22, 2016 at 6:24

Raja Simon

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2 Comments

SAS Over a year ago

Thank you so much! What would the URL be then, please?

SAS Over a year ago

Okay, cool!:D Thank you so much! Your answer was equally helpful! Now I not only can use the code that y'all offered, but also understand what it is doing:)

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