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I started learning django and just deployed my first app when this showed up.
Any suggestions would be useful.
(https://i.stack.imgur.com/GgMjG.png)
it should be
from django.conf.urls import url
(without s at the end)
There is an error in your import, it should be:
from django.conf.urls import url
ref:https://docs.djangoproject.com/en/1.11/ref/urls/
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I am trying to use selenium to login to https://www.amazon.co.uk/ap/signin?showRmrMe=1&openid.return_to=https%3A%2F%2Falexa.amazon.co.uk&openid.identity=http%3A%2F%2Fspecs.openid.net%2Fauth%2F2.0%2Fidentifier_select&openid.assoc_handle=amzn_dp_project_dee_uk&openid.mode=checkid_setup&openid.claimed_id=http%3A%2F%2Fspecs.openid.net%2Fauth%2F2.0%2Fidentifier_select&openid.ns=http%3A%2F%2Fspecs.openid.net%2Fauth%2F2.0&
it works, but then it automatically logs me out whenever I do this automated.
What might cause this issue?
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I have tried looking this up, but for some reason I cannot find anything about it. How do I run a script by giving the particular file directory in start()?
This works (when Test is in the same folder as the main script):
self.process.start("python3 Test.py")
This does NOT work:
self.process.start("python3 /my/path/Test.py")
Try this:
self.process.start("python3 ../my/path/Test.py") # added two periods before "/m"
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I'm currently developing a website with the framework Django (I'm very beginner), but I have a problem with Python: since I have created my templates, I can't run server anymore for this reason (the stack trace points to a line in file urls.py):
<stacktrace>
...
path('apppath/', include('myapp.urls')),
NameError: name 'include' is not defined
Where can I import include from?
Guessing on the basis of whatever little information provided in the question, I think you might have forgotten to add the following import in your urls.py file.
from django.conf.urls import include
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I can't delete the file using os.remove because some problem with the path.
here is the source code: http://www.filedropper.com/install_5
or : http://pastebin.com/L0na3XPm
i would be happy if someone can help me , thank you all
love
I see the line:
rmv=dst+"\\insatll.py"
Is this supposed to be "\\install.py" ?
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When I route back to index with a parameter eg app.route('/index?status=<status>'), why is the URL escaped on the browser side?
http://127.0.0.1:5000/index%3Fstatus%3Dfalse
Or, more towards my question, how do I unescape the URL? And where in Flask am I able to do this? I did some googling around Flask and Werkzeug, but wasn't able to find anything...
Don't put the query string in the route. Instead, use request.args.get to access them.
#app.route('/index')
if 'status' in request.args:
status = request.args.get('status')