How can I get the named parameters from a URL using Flask?

Question

When the user accesses this URL running on my flask app, I want the web service to be able to handle the parameters specified after the question mark:

http://10.1.1.1:5000/login?username=alex&password=pw1

#I just want to be able to manipulate the parameters
@app.route('/login', methods=['GET', 'POST'])
def login():
    username = request.form['username']
    print(username)
    password = request.form['password']
    print(password)

Answer 1

Use request.args to get parsed contents of query string:

from flask import request

@app.route(...)
def login():
    username = request.args.get('username')
    password = request.args.get('password')

Answer 2

The URL parameters are available in request.args, which is a MultiDict that has a get method, with optional parameters for default value (default) and type (type) - which is a callable that converts the input value to the desired format.

from flask import request

@app.route('/my-route')
def my_route():
  page = request.args.get('page', default = 1, type = int)
  filter = request.args.get('filter', default = '*', type = str)

Examples with the code above:

/my-route?page=34               -> page: 34  filter: '*'
/my-route                       -> page:  1  filter: '*'
/my-route?page=10&filter=test   -> page: 10  filter: 'test'
/my-route?page=10&filter=10     -> page: 10  filter: '10'
/my-route?page=*&filter=*       -> page:  1  filter: '*'

Answer 3

You can also use brackets <> on the URL of the view definition and this input will go into your view function arguments

@app.route('/<name>')
def my_view_func(name):
    return name

Answer 4

If you have a single argument passed in the URL you can do it as follows

from flask import request
#url
http://10.1.1.1:5000/login/alex

from flask import request
@app.route('/login/<username>', methods=['GET'])
def login(username):
    print(username)

In case you have multiple parameters:

#url
http://10.1.1.1:5000/login?username=alex&password=pw1

from flask import request
@app.route('/login', methods=['GET'])
    def login():
        username = request.args.get('username')
        print(username)
        password= request.args.get('password')
        print(password)

What you were trying to do works in case of POST requests where parameters are passed as form parameters and do not appear in the URL. In case you are actually developing a login API, it is advisable you use POST request rather than GET and expose the data to the user.

In case of post request, it would work as follows:

#url
http://10.1.1.1:5000/login

HTML snippet:

<form action="http://10.1.1.1:5000/login">
  Username : <input type="text" name="username"><br>
  Password : <input type="password" name="password"><br>
  <input type="submit" value="submit">
</form>

Route:

from flask import request
@app.route('/login', methods=['POST'])
    def login():
        username = request.form.get('username')
        print(username)
        password= request.form.get('password')
        print(password)

Answer 5

url:

http://0.0.0.0:5000/user/name/

code:

@app.route('/user/< string:name >/', methods=['GET', 'POST'])
def user_view(name):
    print(name)

Answer 6

It's really simple. Let me divide this process into two simple steps.

1. On the html template you will declare name tag for username and password as

   <form method="POST">
   <input type="text" name="user_name"></input>
   <input type="text" name="password"></input>
   </form>
   

2. Then, modify your code as:

   from flask import request
   
   @app.route('/my-route', methods=['POST']) #you should always parse username and 
   # password in a POST method not GET
   def my_route():
     username = request.form.get("user_name")
     print(username)
     password = request.form.get("password")
     print(password)
   #now manipulate the username and password variables as you wish
   #Tip: define another method instead of methods=['GET','POST'], if you want to  
   # render the same template with a GET request too