URL
somedomain.com/?_escaped_fragment_
try it:
routes.escaped-fragment.type = "Zend_Controller_Router_Route_Regex"
routes.escaped-fragment.route = "\?_escaped_fragment_"
routes.escaped-fragment.defaults.controller = "index"
routes.escaped-fragment.defaults.action = "someaction"
but he does not see the GET parameter ?_escaped_fragment_
and run
IndexController::indexAction
The ZF router operates on the path only, query string params are stripped off before the route matching occurs, so you're not going to be able to get this working easily. Your options are:
Change your URL structure
Rewrite this URL in .htaccess
Extend/replace the default router to check the query string before doing the standard routing
Related
I have a website building tool created in Django and I'd like to add easy user defined 301 redirects to it.
Webflow has a very easy to understand tool for 301 redirects. You add a path (not just a slug) and then define where that path should lead the user.
I'd like to do the same for the Django project I'm working on. I currently allow users to set a slug that redirects /<slug:redirect_slug>/ and they can set to go to any URL. But I'd like them to be able to add, for example, the path for an old blog post '/2018/04/12/my-favorite-thing/'
What's the best URL conf to use in Django to safely accept any path the user wants?
You can use the Path Converters that convert the path parameters into appropriate types, which also includes a converter for urls.
An example would be like the following:
path('api/<path:encoded_url>/', YourView.as_view()),
As per the docs:
Matches any non-empty string, including the path separator, '/'. This allows you to match against a complete URL path rather than just a segment of a URL path as with str.
In your view, you can get your URL like this:
encoded_url = self.kwargs.get('encoded_url')
Add a RerouteMiddleware which first checks if the request can be served by the existing URLs from the urls.py. If it cannot be served, check if the requested path is from the old -> new URLs mapping, if a match found redirect it to the new URL.
Sample piece of code to try it out.
try:
resolve(request.path_info)
except Resolver404:
# Check if the URL exists in your database/constants
# where you might have stored the old -> new URL mapping.
if request.path is valid:
new_url = # Retrieve the new URL
return redirect(new_url)
response = self.get_response(request)
return response
I am trying to pass a code "Req-2019#000001" Django URL. I want to pass this code as well as normal string and number also in URL as arguments.
path('generateBomForSingleProduct/<requisition_no>/' , views.generateBomForSingleProduct, name='generateBomForSingleProduct'),
Its work properly but the problem is its add extra / before #
My URL is now this
http://127.0.0.1:8000/production/generateBomForSingleProduct/Req-2019/#000001
But I want to like this
http://127.0.0.1:8000/production/generateBomForSingleProduct/Req-2019#000001
Not an extra "/" before my "#"
The portion of the URL which follows the # symbol is not normally sent to the server in the request for the page. So, it's not possible to have a URL as /production/generateBomForSingleProduct/Req-2019#000001
Workaround
Just modify the url as /production/generateBomForSingleProduct/Req-2019/000001, so you need to modify the view also
# views.py
def generateBomForSingleProduct(request, part_1, part_2):
unique_id = "{}#{}".format(part_1, part_2)
# use the "unique_id"
...
#urls.py
urlpatterns = [
...,
path('foo/<part_1>/<part_2>/', generateBomForSingleProduct, name="some-name"),
...
]
Using 'get' would be the proper way to do this. # is for html fragments, which are not sent to the server. Why can't it just be
/production/generateBomForSingleProduct/Req-2019?code=000001 and then you handle everything in the view that way?
Part after # is called fragment identifier, which is used by browsers and never sent to server.
In http://127.0.0.1:8000/production/generateBomForSingleProduct/Req-2019/#000001 url 000001 is never sent to server. Hence using important part of url after # is useless. Better to pass 000001 as query parameters or separate argument.
Suppose my site's domain is mysite.com , now whenever a request comes in this form : mysite.com/https://stackoverflow.com :I want to fetch out this url "https://stackoverflow.com" and send it to the corresponding view.
I have tried this pattern :
url(r'^(?P<preurl>http[s]?://(?:[a-zA-Z]|[0-9]|[$-_#.&+]|[!*(),]|(?:%[0-9a-fA-F][0-9a-fA-F]))+)$',prepend_view)
regex of which matches the incoming appended url and assigns variable preurl the value "https://stackoverflow.com", which I access in corresponding view function .
This works fine for above example but my url pattern is failing in case of some exceptional urls..
Please suggest a robust url pattern by taking into consideration all exceptional urls too, like the following:
ftp://ftp.is.co.za/rfc/rfc1808.txt
http://www.ietf.org/rfc/rfc2396.txt
ldap://[2001:db8::7]/c=GB?objectClass?one
mailto:John.Doe#example.com
news:comp.infosystems.www.servers.unix
tel:+1-816-555-1212
telnet://192.0.2.16:80/
urn:oasis:names:specification:docbook:dtd:xml:4.1.2
That is, if a request comes like :
mysite.com/ldap://[2001:db8::7]/c=GB?objectClass?one
I should be able to get the value "ldap://[2001:db8::7]/c=GB?objectClass?one" in variable preurl
You don't have to make this type of complex url pattern, First, make a URL pattern that matches everything.
url(r'^.*/$', views.fast_track_service, name='fast_track'),
and append it to the end in urlpatterns in your urls.py then in your view, Use request object, So You can get the full path of get request with this method,
fast_track_url = request.get_full_path()[1:]
and then once you got the url try validating that with URLValidator like this.
if not 'http://' in fast_track_url and not 'https://' in fast_track_url:
fast_track_url = 'http://' + fast_track_url
url_validate = URLValidator()
try:
url_validate(fast_track_url)
except:
raise Http404
If you want to validate other complicated URL like mailto etc, then you can write your own validator.
I'm new to node and unable to create a simple route which will include regex as on of the parameter
// student.js - route file for route /student
app.get('/student/:/^[a-z0-9-]+$/', function(req,res){
res.send('student found');
});
when i hit localhost:3000/student/student-slug it says Cannot GET /student/student-slug
two more question
1) how to get param which is of regex, usually we can do this var _student = res.param.student_name but i'm unable to think for the regex
2) how to set optional param, let's say for pagination, route is like
/list/students/ will show list of last x student but /list/students/48 will offset that value to 48th row
this question may be duplicate but i'm unable to find answer
You need to encode the uri string before pass to request and decode it in your route handler.
Usage is very clear:
encodeURIComponent(str);
And for decoding use:
decodeURIComponent(str);
check the official documentation here
also do checkout this blog post on escape vs encode vs encodeURIComponent
Route are defined as
$route['hotels/([a-z]+)/(:any)'] = "hotels/city/$1/$2";
$route['hotels/placename/fivestar'] = "hotels/placename/star/fivestar";
I want to execute the city action in hotels controller when some one type url like
http://website.com/hotels/myspecificcity/fivestar
the above first route is not working. it always load 2nd route . can some one please guide me
i don't want to remove 2nd route though as i will using for other purposes
Thanks