hello ...
your code is good
but you forgot to put comma after second path in urls.py
urlpatterns = [
path('admin/', admin.site.urls),
path('first_app/', include('first_app.urls')),
]
On Saturday, March 23, 2019 at 6:08:24 PM UTC+2, The Aryas wrote:
>
> Hey Guys, I am facing problem o
In your urls.py first_app, change 'index' to ''. That is empty string.
127.0.0.1:8000/first_app
That's all you need. It should work.
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Sorry, it should be first_project
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Great. You are on the right track, only that the Url path you typed in the
address bar triggered the error.
You should use /first_app/index/
Reason:the index url path is located in the first_app. Urls file.
On Sat, Mar 23, 2019, 5:07 PM The Aryas Hey Guys, I am facing problem on django url mappi
Hey Guys, I am facing problem on django url mapping, I did exactly what my
couse said and copied the code text exactly,but it throws the error:-
Page not found (404)
Request Method: GET
Request URL: http://127.0.0.1:8000/index
Using the URLconf defined in protwo.urls, Django tried these URL pat
James - Thank You Very Much
that true it doesn't need /. everything work fine as I want :D.
เมื่อ วันศุกร์ที่ 3 กรกฎาคม ค.ศ. 2015 7 นาฬิกา 38 นาที 54 วินาที UTC+7,
James Schneider เขียนว่า:
>
>
> > > I'm a beginner of Django I want to set my url with database field_name
> > > instead of use prim
> > I'm a beginner of Django I want to set my url with database field_name
> > instead of use primary key from Django tutorial. This is my code.
> >
> > *mysite*
> > **dwru/urls.py**
> > urlpatterns = [
> > url(r'^$', include('product.urls', namespace="product")),
> > ]
>
> This regex is incorr
> I'm a beginner of Django I want to set my url with database field_name
> instead of use primary key from Django tutorial. This is my code.
>
> *mysite*
> **dwru/urls.py**
> urlpatterns = [
> url(r'^$', include('product.urls', namespace="product")),
> ]
This regex is incorrect, and will only
I'm a beginner of Django I want to set my url with database field_name
instead of use primary key from Django tutorial. This is my code.
*mysite***dwru/urls.py**
urlpatterns = [
url(r'^$', include('product.urls', namespace="product")),]
*myapp***product/urls.py**
urlpatterns = [
url(r'^$
Hi
The hash mark has a special meaning in URIs; this is not specific to
Django
If you must use it in an actual URL, it needs to be escaped as %23
The urlencode template tag might help
https://docs.djangoproject.com/en/dev/ref/templates/builtins/#urlencode
Notice that this URL uses a hash mark
Hi Guys,
I have a problem, i have url
like: http://127.0.0.1:8000/get-file?file_path=/home/domagoj/pcap/#1/test.txt
where file_path is path to the file
both
requestGET["file_path"] and
request.get_full_path()
ignore #1/test.txt part, how can i fix this? Is there some other way to get
full ur
On 27.1.2012 4:15, Chen Xu wrote:
When I type the following as the normal python function, it returns the
correct things:
>>> pattern =
re.compile(r'^account_activation\?user_id=(?P[0-9]+)&auth_key=(?P[0-9a-zA-Z]+)$')
>>> match =
pattern.match("account_activation?user_id=36&auth_key=f9717d96aa
When I type the following as the normal python function, it returns the
correct things:
>>> pattern =
re.compile(r'^account_activation\?user_id=(?P[0-9]+)&auth_key=(?P[0-9a-zA-Z]+)$')
>>> match =
pattern.match("account_activation?user_id=36&auth_key=f9717d96aa4e3452a47955225a60862d07119702")
>>> m
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