Pass filepath as parameter to a URL in FLASK(Python)

Question

I want to build an api which accepts a parameter from the user which is a filepath and then process the file given in that path. The file to be processed is already in the server where the api will be running.

As of now, I have written an api where I have hardcoded the filepath in my code which runs the api. Now, I want to configure my api in such a way that accepts a filepath from the user. My api should accept the path as a parameter and process the file that has been given in the path.

The api code is as follows: The convert function returns the category of the file.

import ectd
from ectd import convert
from flask import Flask, request
from flask_restful import Resource, Api
#from flask.views import MethodView

app = Flask(__name__)
api = Api(app)
#convert(r'D:\files\67cecf40-71cf-4fc4-82e1-696ca41a9fba.pdf')    
class ectdtext(Resource):
    def get(self, result):
        return {'data': ectd.convert(result)}

#api.add_resource(ectdtext, '/ectd/<result>')
categories=convert(r'D:\files\6628cb99-a400-4821-8b13-aa4744bd1286.pdf')    
@app.route('/')
def returnResult():
  return categories

if __name__ == '__main__':
     app.run(host="0.0.0.0", port=5000)

So, I want to make changes to this code to accept a parameter from the user which will be a filepath and the convert function will process that filepath. I want to know how to make my api accept a filepath parameter from the user.

Trial with requests.args.get:

    import ectd
    from ectd import convert
    from flask import Flask, request
    from flask_restful import Resource, Api
    #from flask.views import MethodView

    app = Flask(__name__)
    api = Api(app)
    #convert(r'D:\files\67cecf40-71cf-4fc4-82e1-696ca41a9fba.pdf')    
    class ectdtext(Resource):
        def get(self, result):
            return {'data': ectd.convert(result)}

    #api.add_resource(ectdtext, '/ectd/<result>')

    @app.route('/')
    def returnResult():
     categories=convert(r'D:\files\'.format(request.args.get('categories')))
    return categories

    if __name__ == '__main__':
         app.run(host="0.0.0.0", port=5000)

results in error : "RuntimeError: Working outside of request context.

This typically means that you attempted to use functionality that needed an active HTTP request. Consult the documentation on testing for information about how to avoid this problem."

PRESENT SCENARIO:

I am able to post a filepath to the url. My question is now how do I use this posted url with filepath in my code to trigger my function that takes in the filepath and processes the file. Code to post the filepath:

import ectd
from ectd import convert
from flask import Flask, request
from flask_restful import Resource, Api
#from flask.views import MethodView

app = Flask(__name__)
api = Api(app)   
class ectdtext(Resource):
    def get(self, result):
        return {'data': ectd.convert(result)}

#api.add_resource(ectdtext, '/ectd/<result>')
categories=convert('/home/brian/ajay/files/5ca21af9-5b67-45f8-969c-ae571431c665.pdf')   
@app.route('/')
def returnResult():
  return categories



@app.route('/', defaults={'path': ''})
@app.route('/<path:path>')
def get_dir(path):
    return path

#@app.route('/get_dir/<path>')
#def get_dir(path):
#    return path

if __name__ == '__main__':
     app.run(host="0.0.0.0", port=5000)

Pass filepath as parameter to a URL in FLASK(Python)

0 Answers0