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use anki requests client for common user agent name

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some websites are blocking access from the generic requests UA - now
they can choose whether they wish to block Anki or not
This commit is contained in:
Damien Elmes 2018-09-19 23:49:56 +10:00
parent 8f05db5981
commit dfd23f6a41
1 changed files with 4 additions and 1 deletions
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5
aqt/editor.py
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@ -25,6 +25,7 @@ from aqt.utils import shortcut, showInfo, showWarning, getFile, \
import aqt import aqt
from bs4 import BeautifulSoup from bs4 import BeautifulSoup
import requests import requests
from anki.sync import AnkiRequestsClient
pics = ("jpg", "jpeg", "png", "tif", "tiff", "gif", "svg", "webp") pics = ("jpg", "jpeg", "png", "tif", "tiff", "gif", "svg", "webp")
audio = ("wav", "mp3", "ogg", "flac", "mp4", "swf", "mov", "mpeg", "mkv", "m4a", "3gp", "spx", "oga", "webm") audio = ("wav", "mp3", "ogg", "flac", "mp4", "swf", "mov", "mpeg", "mkv", "m4a", "3gp", "spx", "oga", "webm")
@ -658,7 +659,9 @@ to a cloze type first, via Edit>Change Note Type."""))
'User-Agent': 'Mozilla/5.0 (compatible; Anki)'}) 'User-Agent': 'Mozilla/5.0 (compatible; Anki)'})
filecontents = urllib.request.urlopen(req).read() filecontents = urllib.request.urlopen(req).read()
else: else:
r = requests.get(url, timeout=30) reqs = AnkiRequestsClient()
reqs.timeout = 30
r = reqs.get(url)
if r.status_code != 200: if r.status_code != 200:
showWarning(_("Unexpected response code: %s") % r.status_code) showWarning(_("Unexpected response code: %s") % r.status_code)
return return