CODEWITHSUNDEEP
pythonapiflask
Sahil
Sahil

Reputation: 1413

Processing a URL recieved in REST API

I have built an API in Flask that turns HTML pages through their URLS to PDF. However whenever the URL has slashes in it, the script outputs a 404.

this is the code - 

def api(self):
    app = Flask(__name__)
    @app.route('/api/dodo/<filename>/<url>', methods=['GET', 'POST'])

    def convert(url, filename):
        content = request.json
        outfile = pdfkit.from_url(str(url), filename)
        return jsonify({"url":url, 'filename': filename})

    app.run(debug=True)

if the request is localhost/api/dodo/filename/google.com I do get the output file. But if I send this request

http://localhost/api/dodo/filename/https://dev.bizlem.io:8082/scorpio1/GALBRAITH'S_FUEL_27K-55K_ETA_CENTRAL_MED_LIST_-_14TH_JANUARY_2019_1.html

I get 404.

Upvotes: 0

Views: 81

Answers (1)

Vignesh Bayari R.
Vignesh Bayari R.

Reputation: 663

From what I understand, this is happening because the url in the latter case actually maps to a different endpoint. Notice that the slashes added in the url change the meaning.

A better approach is to urlencode filename and url parameters, and pass them as GET parameters (From what I see, /api/dodo is your REST API endpoint and filename, url are parameters).

The code should look somewhat like this:

def api():
    app = Flask(__name__)

    @app.route('/api/dodo', methods=['GET'])
    def convert():
        url = request.args.get('url')
        filename = request.args.get('filename')
        outfile = pdfkit.from_url(url, filename)

        return jsonify({"url":url, 'filename': filename})

    app.run(debug=True)

The url to access the resource would then be:

http://127.0.0.1:5000/api/dodo?url=https%3A%2F%2Fdev.bizlem.io%3A8082%2Fscorpio1%2FGALBRAITH%27S_FUEL_27K-55K_ETA_CENTRAL_MED_LIST_-_14TH_JANUARY_2019_1.html&filename=okay

Upvotes: 1

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