Adding parameter values to a url in flask python

Question

I have the code for the following url:http://localhost/summary/myfile.csv I want the url to look like this:http://localhost/summary?file=myfile.csv

The code is to be written in flask.

My code for the first url is as follows:

@app.route('/summary/<filename>',methods = ['GET'])
def api_summary(filename):
    url = 'C:\\Users\\Desktop\\myproject\\'
    if os.path.exists(url + filename):
        data = pandas.read_csv( url + filename)
        Numeric_Summary = data.describe().to_dict()
        resp = jsonify(Numeric_Summary)
        resp.status_code = 200
        return resp

Nathan Villaescusa · Accepted Answer · 2012-10-30 06:05:00Z

6

You would need to add another route and parse the query string:

from flask import request

@app.route('/summary',methods = ['GET'])
def api_summary_2():
    filename = request.args.get('file', None)
    if filename is None:
        abort(404)
    else:
        return api_summary(filename)

edited Oct 30, 2012 at 6:05

answered Oct 30, 2012 at 5:37

Nathan Villaescusa

17.7k4 gold badges55 silver badges58 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Nathan Villaescusa Over a year ago

How so? Do you get an error? I changed it so that it will accept any filename, not just myfile.csv.

user1403546 Over a year ago

how can you show an input form named 'file' in the web interface and link it to the correct query argument?

Luis Souza Over a year ago

is there a way to insert parameters in request.args (in code, not in the request) ?

Collectives™ on Stack Overflow

Adding parameter values to a url in flask python

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related