I want to see all the routes which my application has. Return them as a response like key=>value pair:
'route1' => '{foo:\w+}'
'route2' => '{baz:\w+\d+}'
... and so on
But I don't know how to get them within my view. For example, this is my view. I want it to return a map of routes. I do this:
#view_config(route_name='route1')
def someView(request):
routes = request.registry.settings.getRoutes() ## what should I print here to get a map of routes?
r = ''
for k,v in sorted(routes.items()):
r += str(k) + "=>" + str(v) + "<br/>";
return Response(r)
There is a RoutesConfiguratorMixin class with get_routes_mapper method. I tried to import the class and called its method but got an error that no registry was in the instance of it:
from pyramid.config.routes import RoutesConfiguratorMixin as Router
r = Router();
routes = r.get_routes_mapper();
## ... and the same code as above
Doesn't work.
There are 2 ways, one is supported (public) and one is unsupported (private).
Option #1 is to use the introspector and is explained here.
Option #2 is to use the route mapper (which is not a public api), in the way that the pyramid debugtoolbar does in its routes panel.
Pyramid installs a bin script called proutes for that purpose.
Install pshell then
pshell to login to pshell with your app config.
then run
print("\n".join([r.path for r in app.routes_mapper.routelist]))
Related
I know this could be a repeated question for many of you but I have not been able to find a proper answer for this yet. I am a beginner to Django and Python. I have a python code which runs and produce output on cli at present but I want the same program to run its output on web.
I read that for web django is best suitable framework and for this purpose I started to study django. I see in every tutorial people have discussed apps, views urls etc but not seen an example which integrate a python code with django.
All I am looking for to understand how can I integrate my python script with Django and where do I place my code in Django project or app. Should I import it within views? if yes, then how to present my output to web.
Here is the sample code I am running, it basically opens two files and run some regex to extract the desired information.
import re
def vipPoolFileOpen(): # function opens vip and pool config file and store them to vip_config and pool_config variables
with open("pool_config.txt",'rb') as pool_config:
pool_config = pool_config.read()
pool_config = pool_config.split('ltm')
with open("vip_config.txt",'rb') as vip_config:
vip_config = vip_config.read()
vip_config = vip_config.split('ltm')
return vip_config,pool_config
def findWidth(vip_config): # function to find the maximum length of vip in entire file, this will be used to adjust column space
colWidth=0
for item in vip_config:
i=0
if colWidth<len(item):
while i<len(vip_config)-1:
if len(item)>=len(vip_config[i+1]):
colWidth=len(item)
i=i+1
else:
i+=1
continue
return colWidth
def regexFunction():
vip_config, pool_config = vipPoolFileOpen()
findWidth(vip_config)
for vip in vip_config:
regVip = re.compile(r'pool (.+)\r')
poolByVip = regVip.findall(vip) # poolByVip holds pool name from the vip_config file
for poolblock in pool_config:
regPool = re.compile(r'pool (.+) {')
poolByConfig = regPool.findall(poolblock)
if poolByVip == poolByConfig:
print vip + poolblock
break
elif poolByVip == ['none']:
print vip
break
else:
continue
Yes, you should present your output to the web via view. You need to write a view function (or a class view) in views.py and provide an url where you want to have it in urls.py
If you rewrite your function to return desired result instead of printing it, tou can do the following:
write this in views.py
from django.http import HttpResponse
from wherever_you_have_it import regexFunction
def bar(request):
result = regexFunction() # result should be a string
return HttpResponse(result)
and in urls.py:
from .views import bar
urlpatterns = [
url(r'^foo$', bar),
]
Providing of course you have created your Django app at the first place.
Your result should be displayed as plain text on address localhost:8000/foo - but you need to:
python menage.py runserver
In your terminal first
And of course feel free to look at:
https://github.com/Ergaro/CheckMyChords
to see how a simple django app looks like
I'm trying to create several URLs on my serv thanks to a loop . The issue is that each function I create in a app.route can't have the same name than the others . And I don't know how to create different function names ...
Here is the code :
json_tweets = []
for line in open('C:\Users\Benjamin\Desktop\DashboardProject\last_rated_set.json',"r"):
json_tweets.append(json.loads(line,"ISO-8859-1"))
cashtag_tab = []
for tweet in json_tweets:
if not(tweet['cashtag'] in cashtag_tab) :
cashtag_tab.append(tweet['cashtag'])
for i in range(0,(len(cashtag_tab)-1)) :
var=cashtag_tab[i]
#app.route("/"+var)
def company(var) :
finance=Share(var)
datas = finance.get_historical('2014-01-01', '2014-12-31')
datas = json.dumps(datas, default=json_util.default)
return datas
I'm getting the error AssertionError : View function mapping is overwritting an existing endpoint function : company
This fails because Flask derives the endpoint name from the function by default, but it would anyway fail later because the function company requires an argument var and the route is not parameterised. The simplest option would be just checking the value inside the handler:
#api.route('/<var>')
def company(var):
if var not in cashtag_tab:
abort(404)
If you want all the routes to be in the routing map for any reason, I once needed a similar thing and came up with something like this:
def url_family(source, methods=('GET',)):
def decorator(f):
for entry in source:
# create a handler that delegates to your function
def view_func(entry=entry, **kwargs):
return f(entry, **kwargs)
endpoint = '{0}_{1}'.format(f.__name__, entry)
url = '/{0}'.format(entry)
api.add_url_rule(url,
methods=methods,
endpoint=endpoint,
view_func=view_func)
return decorator
Then you register the handlers as:
#url_family(cashtag_tab)
def company(var):
...
Assuming that you are using flask now, you should consider Custom URL Converter. Check links below
http://flask.pocoo.org/docs/0.10/api/#flask.Flask.url_map - url_map UrlConverter API
https://exploreflask.com/views.html#url-converters - example url converter
https://stackoverflow.com/a/5872904/3451543 - RegexConverter by Philip Southam
Anyway, specifying more details on your question is always helpful to get accurate answer :)
In flask, I can do this:
render_template("foo.html", messages={'main':'hello'})
And if foo.html contains {{ messages['main'] }}, the page will show hello. But what if there's a route that leads to foo:
#app.route("/foo")
def do_foo():
# do some logic here
return render_template("foo.html")
In this case, the only way to get to foo.html, if I want that logic to happen anyway, is through a redirect:
#app.route("/baz")
def do_baz():
if some_condition:
return render_template("baz.html")
else:
return redirect("/foo", messages={"main":"Condition failed on page baz"})
# above produces TypeError: redirect() got an unexpected keyword argument 'messages'
So, how can I get that messages variable to be passed to the foo route, so that I don't have to just rewrite the same logic code that that route computes before loading it up?
You could pass the messages as explicit URL parameter (appropriately encoded), or store the messages into session (cookie) variable before redirecting and then get the variable before rendering the template. For example:
from flask import session, url_for
def do_baz():
messages = json.dumps({"main":"Condition failed on page baz"})
session['messages'] = messages
return redirect(url_for('.do_foo', messages=messages))
#app.route('/foo')
def do_foo():
messages = request.args['messages'] # counterpart for url_for()
messages = session['messages'] # counterpart for session
return render_template("foo.html", messages=json.loads(messages))
(encoding the session variable might not be necessary, flask may be handling it for you, but can't recall the details)
Or you could probably just use Flask Message Flashing if you just need to show simple messages.
I found that none of the answers here applied to my specific use case, so I thought I would share my solution.
I was looking to redirect an unauthentciated user to public version of an app page with any possible URL params. Example:
/app/4903294/my-great-car?email=coolguy%40gmail.com to
/public/4903294/my-great-car?email=coolguy%40gmail.com
Here's the solution that worked for me.
return redirect(url_for('app.vehicle', vid=vid, year_make_model=year_make_model, **request.args))
Hope this helps someone!
I'm a little confused. "foo.html" is just the name of your template. There's no inherent relationship between the route name "foo" and the template name "foo.html".
To achieve the goal of not rewriting logic code for two different routes, I would just define a function and call that for both routes. I wouldn't use redirect because that actually redirects the client/browser which requires them to load two pages instead of one just to save you some coding time - which seems mean :-P
So maybe:
def super_cool_logic():
# execute common code here
#app.route("/foo")
def do_foo():
# do some logic here
super_cool_logic()
return render_template("foo.html")
#app.route("/baz")
def do_baz():
if some_condition:
return render_template("baz.html")
else:
super_cool_logic()
return render_template("foo.html", messages={"main":"Condition failed on page baz"})
I feel like I'm missing something though and there's a better way to achieve what you're trying to do (I'm not really sure what you're trying to do)
You can however maintain your code and simply pass the variables in it separated by a comma: if you're passing arguments, you should rather use render_template:
#app.route("/baz")
def do_baz():
if some_condition:
return render_template("baz.html")
else:
return render_template("/foo", messages={"main":"Condition failed on page baz"})
I have a complex Flask-based web app. There are lots of separate files with view functions. Their URLs are defined with the #app.route('/...') decorator. Is there a way to get a list of all the routes that have been declared throughout my app? Perhaps there is some method I can call on the app object?
All the routes for an application are stored on app.url_map which is an instance of werkzeug.routing.Map. You can iterate over the Rule instances by using the iter_rules method:
from flask import Flask, url_for
app = Flask(__name__)
def has_no_empty_params(rule):
defaults = rule.defaults if rule.defaults is not None else ()
arguments = rule.arguments if rule.arguments is not None else ()
return len(defaults) >= len(arguments)
#app.route("/site-map")
def site_map():
links = []
for rule in app.url_map.iter_rules():
# Filter out rules we can't navigate to in a browser
# and rules that require parameters
if "GET" in rule.methods and has_no_empty_params(rule):
url = url_for(rule.endpoint, **(rule.defaults or {}))
links.append((url, rule.endpoint))
# links is now a list of url, endpoint tuples
See Display links to new webpages created for a bit more information.
I just met the same question. Those solutions above are too complex.
Just open a new shell under your project:
>>> from app import app
>>> app.url_map
The first 'app' is my project script: app.py,
another is my web's name.
(this solution is for the tiny web with a little route)
I make a helper method on my manage.py:
#manager.command
def list_routes():
import urllib
output = []
for rule in app.url_map.iter_rules():
options = {}
for arg in rule.arguments:
options[arg] = "[{0}]".format(arg)
methods = ','.join(rule.methods)
url = url_for(rule.endpoint, **options)
line = urllib.unquote("{:50s} {:20s} {}".format(rule.endpoint, methods, url))
output.append(line)
for line in sorted(output):
print line
It solves the the missing argument by building a dummy set of options. The output looks like:
CampaignView:edit HEAD,OPTIONS,GET /account/[account_id]/campaigns/[campaign_id]/edit
CampaignView:get HEAD,OPTIONS,GET /account/[account_id]/campaign/[campaign_id]
CampaignView:new HEAD,OPTIONS,GET /account/[account_id]/new
Then to run it:
python manage.py list_routes
For more on manage.py checkout: http://flask-script.readthedocs.org/en/latest/
Apparently, since version 0.11, Flask has a built-in CLI. One of the built-in commands lists the routes:
FLASK_APP='my_project.app' flask routes
Similar to Jonathan's answer I opted to do this instead. I don't see the point of using url_for as it will break if your arguments are not string e.g. float
#manager.command
def list_routes():
import urllib
output = []
for rule in app.url_map.iter_rules():
methods = ','.join(rule.methods)
line = urllib.unquote("{:50s} {:20s} {}".format(rule.endpoint, methods, rule))
output.append(line)
for line in sorted(output):
print(line)
Use cli command in Directory where your flask project is.
flask routes
Since you did not specify that it has to be run command-line, the following could easily be returned in json for a dashboard or other non-command-line interface. The result and the output really shouldn't be commingled from a design perspective anyhow. It's bad program design, even if it is a tiny program. The result below could then be used in a web application, command-line, or anything else that ingests json.
You also didn't specify that you needed to know the python function associated with each route, so this more precisely answers your original question.
I use below to add the output to a monitoring dashboard myself. If you want the available route methods (GET, POST, PUT, etc.), you would need to combine it with other answers above.
Rule's repr() takes care of converting the required arguments in the route.
def list_routes():
routes = []
for rule in app.url_map.iter_rules():
routes.append('%s' % rule)
return routes
The same thing using a list comprehension:
def list_routes():
return ['%s' % rule for rule in app.url_map.iter_rules()]
Sample output:
{
"routes": [
"/endpoint1",
"/nested/service/endpoint2",
"/favicon.ico",
"/static/<path:filename>"
]
}
If you need to access the view functions themselves, then instead of app.url_map, use app.view_functions.
Example script:
from flask import Flask
app = Flask(__name__)
#app.route('/foo/bar')
def route1():
pass
#app.route('/qux/baz')
def route2():
pass
for name, func in app.view_functions.items():
print(name)
print(func)
print()
Output from running the script above:
static
<bound method _PackageBoundObject.send_static_file of <Flask '__main__'>>
route1
<function route1 at 0x128f1b9d8>
route2
<function route2 at 0x128f1ba60>
(Note the inclusion of the "static" route, which is created automatically by Flask.)
You can view all the Routes via flask shell by running the following commands after exporting or setting FLASK_APP environment variable.
flask shell
app.url_map
inside your flask app do:
flask shell
>>> app.url_map
Map([<Rule '/' (OPTIONS, HEAD, GET) -> helloworld>,
<Rule '/static/<filename>' (OPTIONS, HEAD, GET) -> static>])
print(app.url_map)
That, is, if your Flask application name is 'app'.
It's an attribute of the instance of the Flask App.
See https://flask.palletsprojects.com/en/2.1.x/api/#flask.Flask.url_map
Can you tell me how I apply this patch to google app engine as in where to put it? Thank you
def _user_init(self, email=None, _auth_domain=None,
_user_id=None, federated_identity=None, federated_provider=None):
if not _auth_domain:
_auth_domain = os.environ.get('AUTH_DOMAIN')
assert _auth_domain
if email is None and federated_identity is None:
email = os.environ.get('USER_EMAIL', email)
_user_id = os.environ.get('USER_ID', _user_id)
federated_identity = os.environ.get('FEDERATED_IDENTITY',
federated_identity)
federated_provider = os.environ.get('FEDERATED_PROVIDER',
federated_provider)
if not email and not federated_identity:
raise UserNotFoundError
self.__email = email
self.__federated_identity = federated_identity
self.__federated_provider = federated_provider
self.__auth_domain = _auth_domain
self.__user_id = _user_id or None
users.User.__init__ = _user_init
Just use it as-is: Put that code in a module that gets imported before you use the relevant User module or datastore functionality. I included the relevant line to patch the code (the last line) with the patch itself.
Overriding the constructor like this is not safe. If the internal implementation of the Users API changes in production, your application could break without warning.
What are you trying to accomplish here? I don't see any custom logic; it looks like you've just copied the constructor from the SDK verbatim. If you need to add custom logic, try subclassing UserProperty and/or wrapping the users API calls instead.
I think, this belongs to some application as a grep within the appengine sdk, for 'federated_identity' does not result any clues. BTW, you should be doing the same. Grep (or WinGrep) for terms like 'federated' to see if this partial patch can be applied to any source.
Thanks for the updated link. The patch can be added to the file google/appengine/api/users.py
You might just need to add the last line: users.User.__init__ = _user_init
I could figure this out after checking the latest code in the svn.