How to Access the Query String in Flask Routes

How do you access the query string in Flask routes?

from flask import request

@app.route('/data')
def data():
    # here we want to get the value of user (i.e. ?user=some-value)
    user = request.args.get('user')

Capture URL with query string as parameter in Flask route

the path pattern will let you capture more complicated route patterns like URLs:

@app.route('/catch/<path:foo>')
def catch(foo):
    print(foo)
    return foo

The data past the ? indicate it's a query parameter, so they won't be included in that patter. You can either access that part form request.query_string or build it back up from the request.args as mentioned in the comments.

Get a variable from the URL in a Flask route

This is answered in the quickstart of the docs.

You want a variable URL, which you create by adding <name> placeholders in the URL and accepting corresponding name arguments in the view function.

@app.route('/landingpage<id>')  # /landingpageA
def landing_page(id):
    ...

More typically the parts of a URL are separated with /.

@app.route('/landingpage/<id>')  # /landingpage/A
def landing_page(id):
    ...

Use url_for to generate the URLs to the pages.

url_for('landing_page', id='A')
# /landingpage/A

You could also pass the value as part of the query string, and get it from the request, although if it's always required it's better to use the variable like above.

from flask import request

@app.route('/landingpage')
def landing_page():
    id = request.args['id']
    ...

# /landingpage?id=A

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

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')

Flask - how to get query string parameters into the route parameters

You would need to use JavaScript to achieve this so your template would become:

<input type='text' id='address'>
 <button onclick="sendUrl();">submit</button>

<script>
    function sendUrl(){
        window.location.assign("/sales/"+document.getElementById("address").value);
    }
</script>

and your routes similar to before:

@app.route('/sales/')
@app.route('/sales/<address>')
def get_sales(address="Nowhere"):
  # do some magic here
  # render template of sales
  return "The address is "+address

However, this is not the best way of doing this kind of thing. An alternative approach is to have flask serve data and use a single-page-application framework in javascript to deal with the routes from a user interface perspective.

Get query string as function parameters on flask

If you are willing to write a decorator, anything is possible:

from functools import wraps

def extract_args(*names, **names_and_processors):
    user_args = ([{"key": name} for name in names] +
        [{"key": key, "type": processor}
            for (key, processor) in names_and_processors.items()])

    def decorator(f):
        @wraps(f)
        def wrapper(*args, **kwargs):
            final_args, final_kwargs = args_from_request(user_args, args, kwargs)
            return f(*final_args, **final_kwargs)
        return wrapper
    return decorator if len(names) < 1 or not callable(names[0]) else decorator(names[0])

def args_from_request(to_extract, provided_args, provided_kwargs):
    # Ignoring provided_* here - ideally, you'd merge them
    # in whatever way makes the most sense for your application
    results = {}
    for arg in to_extract:
        result[arg["key"]] = request.args.get(**arg)
    return provided_args, results

Usage:

@app.route("/somewhere")
@extract_args("gender", age=int)
def somewhere(gender, age):
    return jsonify(gender=gender, age=age)

How to process GET Query String with Flask

Yes. request.args is a MultiDict:

from flask import Flask, jsonify, request

app = Flask(__name__)

@app.route("/")
def hello():
    all_args = request.args.lists()
    return jsonify(all_args)

Flask-Login | url query string arg 'next'

You need an else statement for the case that _next startswith '/'
In your case _next starts with '/' (or is None) and your given code goes to the return at the bottom:

return render_template('auth/login.html')

using the existing _next as arg inside the url, that is why it sends you to

http://127.0.0.1:5000/auth/login?next=%2Fmedic%2F

So complete the following part of your code with an else statement, othwerise it will send you at the return statement at the bottom (return render_template('auth/login.html')) every time your current condition is not met:

if _next is None or not _next.startswith('/'):
                _next = url_for("medic.home")
                print(_next)
                return redirect(url_for('medic.home'))

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