I have a connexion/flask/werkzeug application and I need to be able to obtain the "base_path" during requests. For example: my application is available at: http://0.0.0.0:8080/v1.0/ui/#/Pet, with the base_path being: "http://0.0.0.0:8080/v1.0".
I want to be able to get the base_path when the requestor performs any defined operation (GET, POST, PUT, etc). I have not been able to find an easy way to obtain the base path. Through a python debugger, I can see the base_path is available higher up in the stack but doesn't appear to be available to the application entrypoint.
#!/usr/bin/env python3
import connexion
import datetime
import logging
from connexion import NoContent
PETS = {}
def get_pet(pet_id):
pet = PETS.get(pet_id)
# >>>--> I WANT TO GET THE BASE_PATH OF THE REQUEST HERE <--<<<
return pet or ('Not found', 404)
For reasons I can not detail due to nda, I have multiple openapi specs for this application and it's important for me to know which base_path is being requested (as they are different). If somebody could help me figure out a way to obtain the base_path per request I would be greatly appreciative :)
Thank you!
Use connexion.request.base_url .
https://connexion.readthedocs.io/en/latest/request.html#header-parameters you can access the connexion.request inside your handler
refer to this topic (Incoming Request Data) from Flask documentation
and dump the incoming request with the before_request hook and extract the right one e.g request.base_url for your case:
from flask import .., request
#bp.before_request
def dump_incoming_request():
from pprint import pprint
pprint(request.__dict__.items())
Related
I would like to know how I can modify the URL to the welcome page.
Currently it is /superset/welcome.
It is run into superset/views/core.py in a #expose('/welcome').
I know I can modify the code inside this #expose, but I want to redirect to another url.
So I want to find the line where there is:
welcome_page = /superset/welcome
As of Superset 1.3, you can change the default landing page by adding this code to your Superset config:
from flask import Flask, redirect
from flask_appbuilder import expose, IndexView
from superset.typing import FlaskResponse
class SupersetDashboardIndexView(IndexView):
#expose("/")
def index(self) -> FlaskResponse:
return redirect("/dashboard/list/")
FAB_INDEX_VIEW = f"{SupersetDashboardIndexView.__module__}.{SupersetDashboardIndexView.__name__}"
In the above example, I am using /dashboard/list/ instead of the default /superset/welcome/.
The code above is Unlicensed and thus is free and unencumbered software released into the public domain.
In superset's file structure, navigate to:
superset/app.py
There you will find
class SupersetIndexView(IndexView):
#expose("/")
def index(self) -> FlaskResponse:
return redirect("/superset/welcome")
Modify this to path where you want to redirect.
I'm using werkzeug in a Django project using Apache/mod_wsgi. What I want to do is access the werkzeug python shell without there actually being an error. The only way I can figure to do this is to intentionally cause an error when the url pattern url(r'^admin/shell', forceAnError()) is matched.
Admittedly, intentionally causing an error isn't the optimal course of action, so if there's a way to simply call/import/render/access the werkzeug python shell from a template or something, that would be the better solution.
If you wrap your WSGI application in a werkzeug.debug.DebuggedApplication with evalex on, you'll get a shell available at /console:
from werkzeug.wrappers import Request, Response
from werkzeug.debug import DebuggedApplication
#Request.application
def app(request):
return Response("Normal application, nothing to see here...")
app = DebuggedApplication(app, evalex=True)
# console_path is another optional keyword argument.
# you can guess what it does.
My python GAE app's central application file looks like this:
import webapp2
import homepage
import user_auth
import user_confirm
import admin_user
import admin_config
import config
app = webapp2.WSGIApplication([
(user_auth.get_login_url(), user_auth.LoginHandler),
(user_auth.get_logout_url(), user_auth.LogoutHandler),
("/user/confirm", user_confirm.UserConfirmHandler),
("/admin/config", admin_config.AdminConfigHandler),
("/admin/user/add", admin_user.AdminAddUserHandler),
("/admin/user", admin_user.AdminUserHandler),
("/", homepage.HomepageHandler),
], debug=True)
As you can see, I must import a bunch of request handlers, but for each request, only one of them is used, the other imports are just useless!
That's a big waste of memory and performance because those unnecessary imports also import other things on their own. Does Google App Engine have some "caching" mechanism or something that makes these unnecessary imports negligible? I think not.
How can I avoid them? I just haven't found out the way to import 1 Request Handler per request. If I put all the routing to app.yaml, that would work the way I want, but it makes things complex because I must write app = webapp2.WSGIApplication(... for every request handler file and repeat those boring urls twice (both in the python file and in app.yaml).
Found the way here, already built into webapp2
http://webapp-improved.appspot.com/guide/routing.html#lazy-handlers
a fews days ago, i tried to learn the python twisted..
and this is how i make my webserver :
from twisted.application import internet, service
from twisted.web import static, server, script
from twisted.web.resource import Resource
import os
class NotFound(Resource):
isLeaf=True
def render(self, request):
return "Sorry... the page you're requesting is not found / forbidden"
class myStaticFile(static.File):
def directoryListing(self):
return self.childNotFound
#root=static.file(os.getcwd()+"/www")
root=myStaticFile(os.getcwd()+"/www")
root.indexNames=['index.py']
root.ignoreExt(".py")
root.processors = {'.py': script.ResourceScript}
root.childNotFound=NotFound()
application = service.Application('web')
sc = service.IServiceCollection(application)
i = internet.TCPServer(8080, server.Site(root))##UndefinedVariable
i.setServiceParent(sc)
in my code, i make an instance class for twisted.web.static.File and override the directoryListing.
so when user try to access my resource folder (http://localhost:8080/resource/ or http://localhost:8080/resource/css), it will return a notFound page.
but he can still open/read the http://localhost:8080/resource/css/style.css.
it works...
what i want to know is.. is this the correct way to do that???
is there another 'perfect' way ?
i was looking for a config that disable directoryListing like root.dirListing=False. but no luck...
Yes, that's a reasonable way to do it. You can also use twisted.web.resource.NoResource or twisted.web.resource.Forbidden instead of defining your own NotFound.
I want to generate for my users temporary download link.
Is that ok if i use django to generate link using url patterns?
Could it be correct way to do that. Because can happen that I don't understand some processes how it works. And it will overflow my memory or something else. Some kind of example or tools will be appreciated. Some nginx, apache modules probably?
So, what i wanna to achieve is to make url pattern which depend on user and time. Decript it end return in view a file.
A simple scheme might be to use a hash digest of username and timestamp:
from datetime import datetime
from hashlib import sha1
user = 'bob'
time = datetime.now().isoformat()
plain = user + '\0' + time
token = sha1(plain)
print token.hexdigest()
"1e2c5078bd0de12a79d1a49255a9bff9737aa4a4"
Next you store that token in a memcache with an expiration time. This way any of your webservers can reach it and the token will auto-expire. Finally add a Django url handler for '^download/.+' where the controller just looks up that token in the memcache to determine if the token is valid. You can even store the filename to be downloaded as the token's value in memcache.
Yes it would be ok to allow django to generate the urls. This being exclusive from handling the urls, with urls.py. Typically you don't want django to handle the serving of files see the static file docs[1] about this, so get the notion of using url patterns out of your head.
What you might want to do is generate a random key using a hash, like md5/sha1. Store the file and the key, datetime it's added in the database, create the download directory in a root directory that's available from your webserver like apache or nginx... suggest nginx), Since it's temporary, you'll want to add a cron job that checks if the time since the url was generated has expired, cleans up the file and removes the db entry. This should be a django command for manage.py
Please note this is example code written just for this and not tested! It may not work the way you were planning on achieving this goal, but it works. If you want the dl to be pw protected also, then look into httpbasic auth. you can generate and remove entries on the fly in a httpd.auth file using htpasswd and the subprocess module when you create the link or at registration time.
import hashlib, random, datetime, os, shutil
# model to hold link info. has these fields: key (charfield), filepath (filepathfield)
# datetime (datetimefield), url (charfield), orgpath (filepathfield of the orignal path
# or a foreignkey to the files model.
from models import MyDlLink
# settings.py for the app
from myapp import settings as myapp_settings
# full path and name of file to dl.
def genUrl(filepath):
# create a onetime salt for randomness
salt = ''.join(['{0}'.format(random.randrange(10) for i in range(10)])
key = hashlib('{0}{1}'.format(salt, filepath).hexdigest()
newpath = os.path.join(myapp_settings.DL_ROOT, key)
shutil.copy2(fname, newpath)
newlink = MyDlink()
newlink.key = key
newlink.date = datetime.datetime.now()
newlink.orgpath = filepath
newlink.newpath = newpath
newlink.url = "{0}/{1}/{2}".format(myapp_settings.DL_URL, key, os.path.basename(fname))
newlink.save()
return newlink
# in commands
def check_url_expired():
maxage = datetime.timedelta(days=7)
now = datetime.datetime.now()
for link in MyDlink.objects.all():
if(now - link.date) > maxage:
os.path.remove(link.newpath)
link.delete()
[1] http://docs.djangoproject.com/en/1.2/howto/static-files/
It sounds like you are suggesting using some kind of dynamic url conf.
Why not forget your concerns by simplifying and setting up a single url that captures a large encoded string that depends on user/time?
(r'^download/(?P<encrypted_id>(.*)/$', 'download_file'), # use your own regexp
def download_file(request, encrypted_id):
decrypted = decrypt(encrypted_id)
_file = get_file(decrypted)
return _file
A lot of sites just use a get param too.
www.example.com/download_file/?09248903483o8a908423028a0df8032
If you are concerned about performance, look at the answers in this post: Having Django serve downloadable files
Where the use of the apache x-sendfile module is highlighted.
Another alternative is to simply redirect to the static file served by whatever means from django.