How can I get the request url in Scrapy's parse()
function? I have a lot of urls in start_urls
and some of them redirect my spider to homepage and as result I have an empty item. So I need something like item['start_url'] = request.url
to store these urls. I'm using the BaseSpider.
The 'response' variable that's passed to parse() has the info you want. You shouldn't need to override anything.
eg. (EDITED)
def parse(self, response):
print "URL: " + response.request.url
The request object is accessible from the response object, therefore you can do the following:
def parse(self, response):
item['start_url'] = response.request.url
You need to override BaseSpider's make_requests_from_url(url)
function to assign the start_url to the item and then use the Request.meta
special keys to pass that item to the parse
function
from scrapy.http import Request
# override method
def make_requests_from_url(self, url):
item = MyItem()
# assign url
item['start_url'] = url
request = Request(url, dont_filter=True)
# set the meta['item'] to use the item in the next call back
request.meta['item'] = item
return request
def parse(self, response):
# access and do something with the item in parse
item = response.meta['item']
item['other_url'] = response.url
return item
Hope that helps.
Instead of storing requested URL's somewhere and also scrapy processed URL's are not in same sequence as provided in start_urls
.
By using below,
response.request.meta['redirect_urls']
will give you the list of redirect happened like ['http://requested_url','https://redirected_url','https://final_redirected_url']
To access first URL from above list, you can use
response.request.meta['redirect_urls'][0]
For more, see doc.scrapy.org mentioned as :
RedirectMiddleware
This middleware handles redirection of requests based on response status.
The urls which the request goes through (while being redirected) can be found in the redirect_urls
Request.meta key.
Hope this helps you
Python 3.5
Scrapy 1.5.0
from scrapy.http import Request
# override method
def start_requests(self):
for url in self.start_urls:
item = {'start_url': url}
request = Request(url, dont_filter=True)
# set the meta['item'] to use the item in the next call back
request.meta['item'] = item
yield request
# use meta variable
def parse(self, response):
url = response.meta['item']['start_url']
来源:https://stackoverflow.com/questions/20081024/scrapy-get-request-url-in-parse