We Keep Coding

Scrapy - Splash fetch dynamic data

I am trying to fetch dynamic phone number from this page (among others): https://www.europages.fr/LEMMERFULLWOOD-GMBH/DEU241700-00101.html
The phone number appears after a click on the element div with the class page-action click-tel. I am trying to get to this data with scrapy_splash using a LUA script to execute a click.
After pulling splash on my ubuntu:
sudo docker run -d -p 8050:8050 scrapinghub/splash
Here is my code so far (I am using a proxy service) :
class company(scrapy.Spider):
name = "company"
custom_settings = {
"FEEDS" : {
'/home/ubuntu/scraping/europages/data/company.json': {
'format': 'jsonlines',
'encoding': 'utf8'
}
},
"DOWNLOADER_MIDDLEWARES" : {
'scrapy_splash.SplashCookiesMiddleware': 723,
'scrapy_splash.SplashMiddleware': 725,
'scrapy.downloadermiddlewares.httpcompression.HttpCompressionMiddleware': 810,
},
"SPLASH_URL" : 'http://127.0.0.1:8050/',
"SPIDER_MIDDLEWARES" : {
'scrapy_splash.SplashDeduplicateArgsMiddleware': 100,
},
"DUPEFILTER_CLASS" : 'scrapy_splash.SplashAwareDupeFilter',
"HTTPCACHE_STORAGE" : 'scrapy_splash.SplashAwareFSCacheStorage'
}
allowed_domains = ['www.europages.fr']
def __init__(self, company_url):
self.company_url = "https://www.europages.fr/LEMMERFULLWOOD-GMBH/DEU241700-00101.html" ##forced
self.item = company_item()
self.script = """
function main(splash)
splash.private_mode_enabled = false
assert(splash:go(splash.args.url))
assert(splash:wait(0.5))
local element = splash:select('.page-action.click-tel')
local bounds = element:bounds()
element:mouse_click{x=bounds.width/2, y=bounds.height/2}
splash:wait(4)
return splash:html()
end
"""
def start_requests(self):
yield scrapy.Request(
url = self.company_url,
callback = self.parse,
dont_filter = True,
meta = {
'splash': {
'endpoint': 'execute',
'url': self.company_url,
'args': {
'lua_source': self.script,
'proxy': 'http://usernamepassword#proxyhost:port',
'html':1,
'iframes':1
}
}
}
)
def parse(self, response):
soup = BeautifulSoup(response.body, "lxml")
print(soup.find('div',{'class','page-action click-tel'}))
The problem is that it has no effect, I still have nothing as if no button were clicked.
Shouldn't the return splash:html() return the results of element:mouse_click{x=bounds.width/2, y=bounds.height/2} (as element:mouse_click() waits for the changes to appear) in response.body ?
Am I missing something here ?

Most times when sites load data dynamically, they do so via background XHR requests to the server. A close examination of the network tab when you click the 'telephone' button, shows that the browser sends an XHR request to the url https://www.europages.fr/InfosTelecomJson.json?uidsid=DEU241700-00101&id=1330. You can emulate the same in your spider and avoid using scrapy splash altogether. See sample implementation below using one url:
import scrapy
from urllib.parse import urlparse
class Company(scrapy.Spider):
name = 'company'
allowed_domains = ['www.europages.fr']
start_urls = ['https://www.europages.fr/LEMMERFULLWOOD-GMBH/DEU241700-00101.html']
def parse(self, response):
# obtain the id and uuid to make xhr request
uuid = urlparse(response.url).path.split('/')[-1].rstrip('.html')
id = response.xpath("//div[#itemprop='telephone']/a/#onclick").re_first(r"event,'(\d+)',")
yield scrapy.Request(f"https://www.europages.fr/InfosTelecomJson.json?uidsid={uuid}&id={id}", callback=self.parse_address)
def parse_address(self, response):
yield response.json()
I get the response
{'digits': '+49 220 69 53 30'}

Wait for Scapy callback function

I am new to Scrapy and Python on general.
Here is the code:
import scrapy
import json
class MOOCSpider(scrapy.Spider):
name = 'mooc'
start_urls = ['https://www.plurk.com/search?q=italy']
custom_settings = {
'DUPEFILTER_CLASS': 'scrapy.dupefilters.BaseDupeFilter',
}
global_id = 1458122036
def parse(self, response):
url = 'https://www.plurk.com/Search/search2'
headers = {
...omitted...
}
for i in range(1,10):
formdata = {
"after_id": str(self.global_id)
}
yield scrapy.FormRequest(url, callback=self.parse_api, formdata=formdata, headers=headers)
def parse_api(self, response):
raw = response.body
data = json.loads(raw)
posts = data["plurks"]
users = data["users"]
l = len(posts)
i = 0
for post in posts:
i = i + 1
if (i == l):
self.global_id = post["plurk_id"]
...omitted code...
yield {
'Author': user_name,
'Body': post['content'],
'app': 'plurk'
}
The problem that I have is that Scrapy is making first all the requests in the for loop and then it is executing the code in parse_api.
What I would like to do is let scrapy do one iteration of the for loop, call the callback function, wait for it to return and then do another iteration.
This because the id that I need for the next request will be set in the global_id variable by the callback function.

You can't achieve this by scheduling requests in loop.
You can implement this only if you will schedule only one (next) request per parse/parse_api method call:
class MOOCSpider(scrapy.Spider):
name = 'mooc'
start_urls = ['https://www.plurk.com/search?q=italy']
custom_settings = {
'DUPEFILTER_CLASS': 'scrapy.dupefilters.BaseDupeFilter',
'DOWNLOAD_DELAY':5,
"USER_AGENT": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/86.0.4240.183 Safari/537.36",
}
def parse(self, response):
# schedule only first request (withour loop)
formdata = {
"query": 'italy',
"start_date": "2019/12",
"end_date": "2020/12",
"after_id": '1458122036', #<- your initial global_id
}
yield scrapy.FormRequest('https://www.plurk.com/Search/search2', callback=self.parse_api, formdata=formdata)
def parse_api(self, response):
data = json.loads(response.body)
after_id = None
for post in data["plurks"]:
after_id = post["plurk_id"]
yield {
'Author': data["users"][str(post["owner_id"])]["nick_name"], # instead of user_id?
'Body': post["content"],
'app': 'plurk'
}
# after end of this loop - after_id should contain required data for next request
# instead of separate loop variable response.meta["depth"] used to limit number requests
if response.meta["depth"] <=11 and after_id: # schedule next request
formdata = {
"query": 'italy',
"start_date": "2019/12",
"end_date": "2020/12",
"after_id": str(after_id),
}
yield scrapy.FormRequest('https://www.plurk.com/Search/search2', callback=self.parse_api, formdata=formdata)

Answering my own question:
Now the parse method does just one request and calls once the parse_api method. Parse_api processes the response and sets the global_id variable. Once it's done processing its own response it makes another request passing itself as the callback function.
By doing this you are guaranteed that the global_id variable will be properly set, since the new request will be made only once parse_api has finished running.
request.cb_kwargs["loop_l"] is used to pass an additional argument to the callback function. This time it's a counter that controls the number of requests we want to make. When the counter is equal to 100 we stop the crawling
import scrapy
import json
plurk_id = []
class MOOCSpider(scrapy.Spider):
name = 'mooc'
start_urls = ['https://www.plurk.com/search?q=']
custom_settings = {
'DUPEFILTER_CLASS': 'scrapy.dupefilters.BaseDupeFilter',
}
global_id = 1455890167
url = 'https://www.plurk.com/Search/search2'
headers = {
...OMITTED...
}
def parse(self, response):
formdata = {
"after_id": str(self.global_id)
}
request = scrapy.FormRequest(self.url, callback=self.parse_api, formdata=formdata, headers=self.headers)
request.cb_kwargs["loop_l"] = str(0)
yield request
def parse_api(self, response, loop_l):
int_loop_l = int(loop_l)
int_loop_l = int_loop_l + 1
if (int_loop_l == 200):
return
raw = response.body
data = json.loads(raw)
...omitted code...
... GET AND SET THE NEW global_id FROM THE RESPONSE ...
# make another request with the new id
formdata = {
"after_id": str(self.global_id)
}
request = scrapy.FormRequest(self.url, callback=self.parse_api, formdata=formdata, headers=self.headers)
request.cb_kwargs["loop_l"] = str(int_loop_l)
yield request

Scrapy-Splash Requested URL vs Real URL

I'm trying to make instantaneous reports for some of my webpages using the Splash and Scrapy-Splash python module.
The problem is that I can't obtain the corect last url like in splash render.json. When a website redirects.
For example on localhost:8050/render.json the result for rendering www.google.com is:
{"requestedUrl": "http://www.google.com/",
"url": "https://www.google.com/?gws_rd=ssl",
"title": "Google", "geometry": [0, 0, 1024, 768]}
But inside my python script I only manage to obtain "http://www.google.com"
My code is:
def start_requests(self):
return [Request(self.url, callback=self.parse, dont_filter=True)]
def parse(self, response):
splash_args = { 'wait': 1 }
return SplashRequest(
response.url,
self.parse_link,
args=splash_args,
endpoint='render.json',
)
def parse_link(self, response):
result = {
'requested_url': response.data['requestedUrl'],
'real_url': response.data['url'],
'response': response.request.url,
'splash_url': response.real_url
}
But any of this returns:
{"requested_url": "http://www.google.com/",
"real_url": "http://www.google.com/",
"response": "http://127.0.0.1:8050/render.json",
"splash_url": "http://127.0.0.1:8050/render.json"}
Is there any way to overcome this?

Python - Scrapy splash can't render this page

https://www.miamidade.realforeclose.com/index.cfm?zaction=AUCTION&Zmethod=PREVIEW&AUCTIONDATE=08/16/2018
This is the page that i'm trying to scrape. When i use SplashRequest to open it i get a different page with the same source.
Those are my settings for splas:
ROBOTSTXT_OBEY = False
SPLASH_URL = 'http://192.168.99.100:8050'
DOWNLOADER_MIDDLEWARES = {
'scrapy_splash.SplashCookiesMiddleware': 723,
'scrapy_splash.SplashMiddleware': 725,
'scrapy.downloadermiddlewares.httpcompression.HttpCompressionMiddleware':
810,
}
SPIDER_MIDDLEWARES = {
'scrapy_splash.SplashDeduplicateArgsMiddleware': 100,
}
DUPEFILTER_CLASS = 'scrapy_splash.SplashAwareDupeFilter'
My spider code:
import scrapy
from scrapy_splash import SplashRequest
class RealForeclosure(scrapy.Spider):
name = 'realForeclosure'
start_urls = [
'https://www.miamidade.realforeclose.com/index.cfm?
zaction=user&zmethod=calendar'
]
def parse(self,response):
link = 'https://www.miamidade.realforeclose.com/index.cfm?
zaction=AUCTION&Zmethod=PREVIEW&AUCTIONDATE='
date = response.xpath('//div[#tabindex="0"]/#dayid').extract()[10]
yield SplashRequest(link+date, callback=self.auction)
def auction(self, response):
for i in response.css('.AUCTION_ITEM').extract():
yield {'item':i}

You need some kind of delay to allow Splash render result:
script1 = """
function main(splash, args)
assert (splash:go(args.url))
assert (splash:wait(0.5))
return {
html = splash: html(),
png = splash:png(),
har = splash:har(),
}
end
"""
def parse(self,response):
link = 'https://www.miamidade.realforeclose.com/index.cfm?zaction=AUCTION&Zmethod=PREVIEW&AUCTIONDATE='
date = response.xpath('//div[#tabindex="0"]/#dayid').extract()[10]
yield SplashRequest(
link+date,
callback=self.auction,
endpoint='execute',
args={
'html': 1,
'lua_source': self.script1,
'wait': 0.5,
}
)

Scrapy: How to get cookies from splash

I am trying to get the cookies from a splash request, but I keep getting an error.
Here is the code I am using:
class P2PEye(scrapy.Spider):
name = 'p2peyeSpider'
allowed_domains = ['p2peye.com']
start_urls = ['https://www.p2peye.com/platform/h9/']
def start_requests(self):
script = '''
function main(splash)
local url = splash.args.url
assert(splash:go(url))
assert(splash:wait(0.5))
return {
cookies = splash:get_cookies(),
}
end
'''
for url in self.start_urls:
yield SplashRequest(url, callback=self.parse, endpoint='render.html',args={'wait': 1, 'lua_source': script})
def parse(self, response):
print(response.request.headers.getlist('Set-Cookie'))
print(response.cookiejar)
This is my settings.py
SPLASH_URL = 'http://127.0.0.1:8050'
CRAWLERA_ENABLED= False
DOWNLOADER_MIDDLEWARES = {
'scrapy_splash.SplashCookiesMiddleware': 723,
'scrapy_splash.SplashMiddleware': 725,
'scrapy.downloadermiddlewares.httpcompression.HttpCompressionMiddleware': 810,
}
SPIDER_MIDDLEWARES = {'scrapy_splash.SplashDeduplicateArgsMiddleware': 100 }
DUPEFILTER_CLASS = 'scrapy_splash.SplashAwareDupeFilter'
HTTPCACHE_STORAGE = 'scrapy_splash.SplashAwareFSCacheStorage'
COOKIES_ENABLED = True
COOKIES_DEBUG = True
SPLASH_COOKIES_DEBUG = True
The result of response.request.headers.getlist('Set-Cookie') is [],
and response.cookiejar got an error: AttributeError: 'SplashTextResponse' object has no attribute 'cookiejar'.
So how can I get the cookies without causing an error?

To access response.cookiejar you need to return SplashJsonResponse
try returning extra fields on your Lua script:
script = '''
function main(splash)
local url = splash.args.url
assert(splash:go(url))
assert(splash:wait(0.5))
local entries = splash:history()
local last_response = entries[#entries].response
return {
url = splash:url(),
headers = last_response.headers,
http_status = last_response.status,
cookies = splash:get_cookies(),
html = splash:html(),
}
end
'''

Using the LUA script below the response will be a dict with cookies located at key cookies
function main(splash)
local url = splash.args.url
assert(splash:go(url))
assert(splash:wait(0.5))
return {
cookies = splash:get_cookies(),
}
end
So to access you should use
# d = requests.post('splash').json()
print(d['cookies'])