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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')
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I use this library: https://github.com/yusufusta/ytstudio
When you are uploading video, to interact with youtube api you need to have SESSION_TOKEN
I checked documentation and i have seen that i need to make request to studio.youtube.com and get response of /grst endpoint.
Can i get SESSION_TOKEN without entering to youtube studio in python automatically?
I tried to send request to /grst but i couldn't do it.
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My problem:
I'm using beautiful SOAP in Python, and i want to know how do i get the concrete attribute such as "data-hk".
My code at the moment:
The output of the code is km/L, but i want the data about HK. How do i specifically select the right attribute within the span?
Many thanks in advance.
I tried the above code, and I've stated the result and output of it above.
Try this:
HK = cars.find("span", class_="variableDataColumn")["data-hk"]
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Which of the following regular expressions can be used to get the domain name?
I try the next code but it doesn't work, there is something that i'm doing wrong?
In the picture the another options
txt = 'I refer to https://google.com and i never refer http://www.baidu.com'
print(txt.findall(?<=https:\/\/)([A-Za-z0-9.]*))
You selected the correct regexp, you just have to quote it to use it in Python. You also need to call re.findall(), it's not a string method.
import re
txt = 'I refer to https://google.com and i never refer http://www.baidu.com'
print(re.findall(r'(?<=https:\/\/)([A-Za-z0-9.]*)', txt))
Here's a regex that'll get your URLs
http(s?)://(?:[a-z0-9](?:[a-z0-9-]{0,61}[a-z0-9])?\.)+[a-z0-9][a-z0-9-]{0,61}[a-z0-9]
It'll work for https://stackoverflow.com, http://example.com, https://example.com etc...
If you don't want the http or https just use this:
(?:[a-z0-9](?:[a-z0-9-]{0,61}[a-z0-9])?\.)+[a-z0-9][a-z0-9-]{0,61}[a-z0-9]
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I have created a flask application that stores blog posts inside mongodb. But Now I face a problem when I attempt to show images like <img src="/static/img/blog.jpg">inside those posts. Because those double quotes create problem in mongo db. When it appears in html it looks something like this <img src = "/static/img/blog.jpg">
please help
Whenever you are passing double quotes in a string you would need to escape the characters.
"<Img src=\"url\">"
Also in jinja if you want to render the html code you need to use |safe
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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/