I have a script called output.py
This script takes in 2 inputs, fileA and file B.
I can run it on my terminal by using the command output.py -fileA -fileB. The script will create a new JSON file and save it to the directory.
I want to run this script using Flask. I've defined a bare bones App here but I'm not sure how I'd run this using Flask
from flask import Flask
import output
import scripting
app = Flask(__name__)
#app.route('/')
def script():
return output
if __name__ == '__main__':
app.run()
Can someone help me out here, thanks!
It appears you are new to Flask. Get some basic tutorials (there are many on the web).
There are a couple of options:
Send contents of file A and File b as json payload. Pull the A and B content off the json body and do the processing you need to and return the body.
Send the contents of the file as multipart/form-data (you can send multiple files).
Note: This is not working code - just for illustration.
from flask import Flask, request, make_response
app = Flask(__name__)
def build_response(status=False, error="", data={}, total=0, headers=[], contentType="application/json", expose_headers=["X-Total-Count"], retcode=400, additional_data=None):
resp = {"success": status, "error": error, "data": data}
resp = make_response(json.dumps(resp))
for item in headers:
resp.headers[item] = headers[item]
resp.headers['Content-Type'] = contentType
resp.headers.add('Access-Control-Expose-Headers', ','.join(expose_headers))
resp.status_code = retcode
return resp
#app.route('/run-script', methods=['POST'])
def run_script():
# check if the post request has the file part
try:
# Note: THis code is just to illustrate the concept.
# Option-1 (content type must be application/json)
json_dict = request.get_json()
fileA = json_dict["fileA"]
fileB = json_dict["fileB"]
# Option-2 (Note: fileA/fileB are objects, put a pdb and check it out)
fileA = request.files['fileA']
fileB = request.files['fileA']
resp = process(fileA, fileB)
return build_response(status=True, data=resp, retcode=200)
except Exception as e:
msg = f"Error - {str(ec)}"
return build_response(status=False, error=msg, retcode=400)
Related
So, in my main app.py, I have :
app = create_app()
...
app.run()
In __init__.py I have:
def create_app():
...
app = Flask(__name__)
setup_logging(app)
return app
def setup_logging(app):
RequestID(app)
handler = logging.FileHandler(app.container.config.get("app.LOG_FILE"))
handler.setFormatter(logging.Formatter("%(asctime)s : %(levelname)s : %(request_id)s - %(message)s"))
handler.addFilter(RequestIDLogFilter()) # << Add request id contextual filter
logging.getLogger().addHandler(handler)
logging.getLogger().setLevel(level="DEBUG")
And in my routes.py, I have: (This wiring/linking is done with the help of dependency injection)
def configure(app):
app.add_url_rule("/", "check", check)
...
def check():
logging.info("You hit /")
# Here I want to return the Request UUID that's generated by RequestLogIDFilter
return make_response(jsonify(dict(ok=True)), 200)
How do I access the Request UUID generated by the RequestLogIDFilter? In my log file I correctly see the log messages as:
2022-08-15 07:00:18,030 : INFO : 27b437fd-98be-4bc9-a609-912043e3a38e - Log message test memberId=9876
I want to take this value 27b437fd-98be-4bc9-a609-912043e3a38e and include it in the response headers as X-REQUEST-UUID: 27b437fd-98be-4bc9-a609-912043e3a38e
An alternative was to rip out RequestLogIDFilter and only work with flask-request-id-header except, there's no way to write it to log file (logging.FileHandler does not have a write() method so it fails)
This is something trivial I'm sure but the official docs nowhere mention how to access these request ids for logging to file or sending back as a response header.
If you are using flask_log_request_id then you probably want this as given in their github repo
Code copied here in case link does not work
from flask_log_request_id import current_request_id
#app.after_request
def append_request_id(response):
response.headers.add('X-REQUEST-ID', current_request_id())
return response
I have created a simple flask app that and I'm reading the response from python as:
response = requests.post(url,data=json.dumps(data), headers=headers )
data = json.loads(response.text)
Now my issue is that under certain conditions I want to return a 400 or 500 message response. So far I'm doing it like this:
abort(400, 'Record not found')
#or
abort(500, 'Some error...')
This does print the message on the terminal:
But in the API response I kept getting a 500 error response:
The structure of the code is as follows:
|--my_app
|--server.py
|--main.py
|--swagger.yml
Where server.py has this code:
from flask import render_template
import connexion
# Create the application instance
app = connexion.App(__name__, specification_dir="./")
# read the swagger.yml file to configure the endpoints
app.add_api("swagger.yml")
# Create a URL route in our application for "/"
#app.route("/")
def home():
"""
This function just responds to the browser URL
localhost:5000/
:return: the rendered template "home.html"
"""
return render_template("home.html")
if __name__ == "__main__":
app.run(host="0.0.0.0", port="33")
And main.py has all the function I'm using for the API endpoints.
E.G:
def my_funct():
abort(400, 'Record not found')
When my_funct is called, I get the Record not found printed on the terminal, but not in the response from the API itself, where I always get the 500 message error.
You have a variety of options:
The most basic:
#app.route('/')
def index():
return "Record not found", 400
If you want to access the headers, you can grab the response object:
#app.route('/')
def index():
resp = make_response("Record not found", 400)
resp.headers['X-Something'] = 'A value'
return resp
Or you can make it more explicit, and not just return a number, but return a status code object
from flask_api import status
#app.route('/')
def index():
return "Record not found", status.HTTP_400_BAD_REQUEST
Further reading:
You can read more about the first two here: About Responses (Flask quickstart)
And the third here: Status codes (Flask API Guide)
I like to use the flask.Response class:
from flask import Response
#app.route("/")
def index():
return Response(
"The response body goes here",
status=400,
)
flask.abort is a wrapper around werkzeug.exceptions.abort which is really just a helper method to make it easier to raise HTTP exceptions. That's fine in most cases, but for restful APIs, I think it may be better to be explicit with return responses.
Here's some snippets from a Flask app I wrote years ago. It has an example of a 400 response
import werkzeug
from flask import Flask, Response, json
from flask_restplus import reqparse, Api, Resource, abort
from flask_restful import request
from flask_cors import CORS
app = Flask(__name__)
CORS(app)
api = Api(app)
parser = reqparse.RequestParser()
parser.add_argument('address_to_score', type=werkzeug.datastructures.FileStorage, location='files')
class MissingColumnException(Exception):
pass
class InvalidDateFormatException(Exception):
pass
#api.route('/project')
class Project(Resource):
#api.expect(parser)
#api.response(200, 'Success')
#api.response(400, 'Validation Error')
def post(self):
"""
Takes in an excel file of addresses and outputs a JSON with scores and rankings.
"""
try:
df, input_trees, needed_zones = data.parse_incoming_file(request)
except MissingColumnException as e:
abort(400, 'Excel File Missing Mandatory Column(s):', columns=str(e))
except Exception as e:
abort(400, str(e))
project_trees = data.load_needed_trees(needed_zones, settings['directories']['current_tree_folder'])
df = data.multiprocess_query(df, input_trees, project_trees)
df = data.score_locations(df)
df = data.rank_locations(df)
df = data.replace_null(df)
output_file = df.to_dict('index')
resp = Response(json.dumps(output_file), mimetype='application/json')
resp.status_code = 200
return resp
#api.route('/project/health')
class ProjectHealth(Resource):
#api.response(200, 'Success')
def get(self):
"""
Returns the status of the server if it's still running.
"""
resp = Response(json.dumps('OK'), mimetype='application/json')
resp.status_code = 200
return resp
You can return a tuple with the second element being the status (either 400 or 500).
from flask import Flask
app = Flask(__name__)
#app.route('/')
def hello():
return "Record not found", 400
if __name__ == '__main__':
app.run()
Example of calling the API from python:
import requests
response = requests.get('http://127.0.0.1:5000/')
response.text
# 'This is a bad request!'
response.status_code
# 400
I think you're using the abort() function correctly. I suspect the issue here is that an error handler is that is catching the 400 error and then erroring out which causes the 500 error. See here for more info on flask error handling.
As an example, the following would change a 400 into a 500 error:
#app.errorhandler(400)
def handle_400_error(e):
raise Exception("Unhandled Exception")
If you're not doing any error handling, it could be coming from the connexion framework, although I'm not familiar with this framework.
You can simply use #app.errorhandler decorator.
example:
#app.errorhandler(400)
def your_function():
return 'your custom text', 400
Hi I am new to writing web APIs in python. And my understanding of REST is limited
I have a simple Flask API that takes in a python dict {'pdf':pdf_as_bytes, 'filename':string}
The below is my server script:
#app.route("/predict", methods=["POST"])
def predict():
data = {"success": False}
if flask.request.method == "POST":
pdf = flask.request.files["pdf"].read()
filename = flask.request.files["filename"].read().decode("utf-8")
assert isinstance(filename, str), "Got {}".format(type(filename))
assert isinstance(pdf, bytes), "Got {}".format(type(pdf))
# further processing happens and returns a json
This works as intended when I write a python client as follows:
import requests
import os
ip = "localhost"
port = 8605
url = "http://{}:{}/predict".format(ip,port)
path_to_pdf = "./617339931.pdf"
with open(path_to_pdf, "rb") as f:
pdf = f.read() # pdf is a bytes
# the payload must have the following fields: "pdf": bytes, "filename": string object
payload = {"pdf":pdf,"filename":os.path.basename(path_to_pdf).split(".")[0]}
# post request
result = requests.post(url=url, files=payload).json()
# the resulting json always has a field called as success, which returns True or False accordingly
if result["success"] == True:
print(result["data"].keys())
But, When I send a request using Postman I get a 400 Error! Below is the screen shot of the error
I don't understand. How can I change my server code so that it works with Postman and also Python client programs
I just did the same thing, and I think it's because of the double quotes you are putting in key and value, try to take them out.
I have a flask app that was created using a csv file. Using html forms, I give the user the option to input new data, and then I write changes to the csv (using a custom function) and download it. This downloads just fine and saves to my desktop. Is there any way to tweak the code to save it in the same project directory and overwrite the csv that serves the flask app? This way the app might update upon refresh. Thanks!
#app.route('/csv/')
def download_csv():
model_id=request.args['textid']
client_id = session['client_id']
# return response
df=recommender.update_history(client_id, model_id)
df= recommender.get_csv()
resp = make_response(df.to_csv(encoding='iso-8859-1',index=False))
resp.headers["Content-Disposition"] = "attachment; filename=export.csv"
resp.headers["Content-Type"] = "text/csv"
return resp
The comment above is correct. I didn't need the make_response function. Here's what worked:
#app.route('/csv/')
def download_csv():
model_id=request.args['textid']
client_id = session['client_id']
# return response
df=recommender.update_history(client_id, model_id)
df= recommender.get_csv()
path=r'data/file_name.csv'
resp = df.to_csv(path, encoding='iso-8859-1',index=False)
return render_template('index.html')
I lately started using Flask in one of my projects to provide data via a simple route. So far I return a json file containing the data and some other information. When running my Flask app I see the status code of this request in terminal. I would like to return the status code as a part of my final json file. Is it possible to catch the same code I see in terminal?
Some simple might look like this
from flask import Flask
from flask import jsonify
app = Flask(__name__)
#app.route('/test/<int1>/<int2>/')
def test(int1,int2):
int_sum = int1 + int2
return jsonify({"result":int_sum})
if __name__ == '__main__':
app.run(port=8082)
And in terminal I get:
You are who set the response code (by default 200 on success response), you can't catch this value before the response is emited. But if you know the result of your operation you can put it on the final json.
#app.route('/test/<int1>/<int2>/')
def test(int1, int2):
int_sum = int1 + int2
response_data = {
"result": int_sum,
"sucess": True,
"status_code": 200
}
# make sure the status_code on your json and on the return match.
return jsonify(response_data), 200 # <- the status_code displayed code on console
By the way if you access this endpoint from a request library, on the response object you can find the status_code and all the http refered data plus the json you need.
Python requests library example
import requests
req = requests.get('your.domain/test/3/3')
print req.url # your.domain/test/3/3
print req.status_code # 200
print req.json() # {u'result': 6, u'status_code: 200, u'success': True}
You can send HTTP status code as follow:
#app.route('/test')
def test():
status_code = 200
return jsonify({'name': 'Nabin Khadka'}, status_code) # Notice second element of the return tuple(return)
This way you can control what status code to return to the client (typically to web browser.)