It´s muy first time developing an API. It's very simple. here is my code:
from flask import Flask, Response
from flask import request
from flask import jsonify
import pyodbc
from sqlalchemy import create_engine,Integer
import pandas as pd
import urllib
app = Flask(__name__)
params = urllib.parse.quote_plus("DRIVER={SQL Server Native Client 11.0};"
"SERVER=xxx.xx.x.x;"
"DATABASE=xxxxx;"
"UID=xxxx;"
"PWD=xxxx")
engine = create_engine('mssql+pymssql://xxxx:xxxx#xxx.xx.x.x/xxxx')
sql_talle = "RSCV_TALLE_PARA_CADA_MARCA.sql"
fd1 = open(sql_talle, 'r')
sqlFile1 = fd1.read()
fd1.close()
DF1 = pd.read_sql_query(sqlFile1,engine)
DF1.insert(14, 'PROVEEDOR', DF1.pop('PROVEEDOR'))
periodo=DF1['YEARMONTH_DATE'].drop_duplicates()
NIKE=DF1.loc[DF1['MARCA'] == 'NIKE']
NIKE=NIKE.to_dict('records')
#app.route('/NIKE',methods = ['GET'])
def show_codigo_Agrupador():
if request.method == 'GET':
response = jsonify({'NIKE':NIKE})
response.status_code = 200
return response
if __name__ == "__main__":
app.run(debug=True)
it´s working fine!
when I run the .py on a terminal I get that it´s Running on http://127.0.0.1:5000
After that I try to do a request on another .py, simulating an external user:
import requests
import json
url = 'http://127.0.0.1:5000/NIKE'
r=requests.get(url)
response=r.text
j=response.json()
print(j)
but I get this error:
Traceback (most recent call last):
File "c:\Users\mvazquez\DABRA\FLASK_API\prueba_request_api.py", line 7, in <module>
j=response.json()
AttributeError: 'str' object has no attribute 'json'
I have these questions:
what am I doing wrong here?
my data is based on an sql query, I need to run the script every day to have refresehed data or it is refreshed when request is done? (sorry if it's stupid, I have no idea about API)
thanks in advance!
Related
I have developed a flask application that returns some text from OPEN-AI by giving some inputs.
But unfortunately the rest API in my application returns 404 error.
Here is the code:
from crypt import methods
from warnings import catch_warnings
from flask import Flask,request
from flask_cors import CORS
import flask
import openai
from flask_restful import Api,Resource
import base64
import json
#Init
app = Flask(__name__)
CORS(app)
api = Api(app)
app.run(host='0.0.0.0',port=8080)
#OPENAI CREDENTIALS
openai.api_key = ""
#Functions
class advert(Resource):
def post(self):
try:
request_body=request.json
A=request_body["data"]
adprompt = "Write an advertisement for " + A
response = openai.Completion.create(
engine="text-davinci-002",
prompt=adprompt,
temperature=0.7,
max_tokens=70,
top_p=1.0,
n=1
)
json_advert = json.loads(str(response))
advert_output = json_advert['choices'][0]['text']
to_return= json_advert = json.loads(str(advert_output))
return to_return,200
except:
return ({"ERROR":"Error Occured"}),500
#Mapping
api.add_resource(advert,'/data',methods=['POST'])
if __name__=='__main__':
app.run(debug=True)
This is the response i get:
192.168.1.21 - - [24/Nov/2022 11:52:59] "POST /data HTTP/1.1" 404 -
I've tried changing the port and endpoints, nothing helped .
How to solve this.
Your problem is at this line,
app.run(host='0.0.0.0',port=8080)
take it out, then add the parameters into the last line,
if __name__=='__main__':
app.run(debug=True, host='0.0.0.0',port=8080)
A bit more explanation, when the code hits app.run, it actually starts the server and not running the line api.add_resource. Hence, no path was configured.
I am new to flask and I want to return 401 error, I am not using authentication, I am simply using a parameter api_key to check whether the user is authenticated.
I've tried abort(401) and return false, but it returns error 400 'Bad Request', is this the default way it returns a 401. Or is there a different way to do this?
My Code:
import flask
from flask import jsonify, request, Response
from flask_cors import CORS, cross_origin
from GoogleNews import GoogleNews
googlenews = GoogleNews()
googlenews.set_lang('en')
googlenews.set_period('1d')
googlenews.set_encode('utf-8')
app=flask.Flask(__name__)
cors = CORS(app)
app.config['CORS_HEADERS'] = 'Content-Type'
#cross_origin()
#app.route('/', methods=["GET"])
def return_news():
API_KEY = request.args['api_key']
if API_KEY == "xyz":
googlenews.get_news('TECHNOLOGY')
return jsonify(googlenews.result(True)[0:4])
else:
return Response(response="Unauthorized", status=401)
Thankyou
instead, try flask.Response object.
from flask import Flask, Response
app = Flask(__name__)
#app.route("/abort")
def abort():
return Response(response="Unauthorized", status=401)
app.run()
without the actual code it is hard to tell, but I think it is not using the flask abort() but the python internal abort() function.
Check the code below:
import flask
app = flask.Flask("aborter")
#app.route("/abort")
def abort():
return flask.abort(401)
if __name__ == '__main__':
app.run()
Happy Coding
I am trying to write a lambda function in 3.8 version but I am getting error while doing a get requests
[ERROR] AttributeError: module 'botocore.vendored.requests' has no attribute 'get' Traceback (most recent call last): File "/var/task/lambda_function.py"
import json
from botocore.vendored import requests
def lambda_handler(event, context):
request = event['Records'][0]['cf']['request']
print (request)
print(request['headers'])
token = request['headers']['cookie'][0]['value'].partition("=")[2]
print (token)
print(type(request['uri']))
consumer_id = request['uri'].rpartition('/')[-1]
print (consumer_id)
#Take the token and send it somewhere
token_response = requests.get(url = 'https://url/api/files/' + consumer_id, params = {'token': token})
print (token_response)
return request
I tried following this blog https://aws.amazon.com/blogs/compute/upcoming-changes-to-the-python-sdk-in-aws-lambda/
but not able to identify which layer to add. Could anyone please help
According to the link you provided and assuming that request was correctly installed you should be using
import requests
instead of
from botocore.vendored import requests
I'm using Python 3.7.4 and http.server module to receive POST requests that contain a file from an ERP.
Everything works fine (file get received and written correctly) except ERP get connection timeout error.
It's first time I use http.server which looks pretty simple but for sure I'm missing something.
See code below, isnt't "self.send_response(200)" enough?
On ERP vendor website they provide an example in PHP to receive data:
if (move_uploaded_file ($_FILES['file']['tmp_name'], "items.xml")){
echo "OK";
} else {
echo "Error";
}
So ERP expect "OK" after successful connection/transfer
Here it is my Python code:
from http.server import BaseHTTPRequestHandler, HTTPServer
import time
import cgi
class Test_Server(BaseHTTPRequestHandler):
def do_POST(self):
print("POST request received")
self.send_response(200)
form = cgi.FieldStorage(fp=self.rfile, headers=self.headers, environ={'REQUEST_METHOD': 'POST', 'CONTENT_TYPE': self.headers['Content-Type'],})
f = open("data/test-orig.xml","wb+")
f.write(form['file'].value)
f.close()
httpd = HTTPServer((hostName, hostPort), Test_Server)
print(time.asctime(), "Server Starts - %s:%s" % (hostName, hostPort))
try:
httpd.serve_forever()
except KeyboardInterrupt:
pass
httpd.server_close()
print(time.asctime(), "Server Stops - %s:%s" % (hostName, hostPort))
Best regards,
cid
Manage to do it easily with Flask:
from flask import Flask
from flask import request
app = Flask(__name__)
#app.route('/post-data', methods=['POST'])
def test_server():
data = request.files['file']
data.save('data/test.xml')
return "OK"
if __name__ == '__main__':
app.run(host='0.0.0.0')
Solved!
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)