I am using Scrapy and am having trouble with the script. It works fine with the shell:
scrapy shell "www.redacted.com" I use response.xpath("//li[#a data-urltype()"]).extract
I am able to scrape 200 or so links from the page.
Here is the code from the webpage I am trying to scrape:
<a data-urltype="/view" data-mce-href="http://www.redacted.aspx?ID=xxxxxxxxxx" data-linktype="external" href="http://www.redacted.com/Home/wfContent.aspx?xxxxxxxxxxxxx" data-val="http://www.redacted.gov/Home/wfContent.aspx?xxxxxxxxxxxx" target="_blank">link text</a>
My problem is the script: (posted below) I know the "a data-val" is wrong.
import scrapy
from ..items import LinkscrapeItem
class Linkscrape(scrapy.Spider):
name = 'lnkscrapespider'
start_urls = [
'https://www.redacted.com'
]
def parse(self, response):
items = LinkscrapeItem()
links = response.xpath("a data-val").xpath.extract()
for links in links:
items['links'] = links
yield{
'links': links
}
You don't need to use .xpath() twice:
links = response.xpath("//li/a/#data-val").extract()
# or
links = response.xpath("//li/a/#data-val").getall()
Also below doesn't make sense (may be you need for link in links? ):
for links in links:
items['links'] = links
yield{
'links': links
}
if you are going to scrape data-val from a. use below xpath.
links = response.xpath("//li/a/#data-val").xpath.extract()
Related
I want help in scraping infinite scrolling pages. For now, I have entered pageNumber = 100, which helps me in getting the name from 100 pages.
But I want to crawl all the pages till the end. As the page has infinite scrolling and being new to scrapy I am unable to do the same. I am trying this for the past 2 days.
class StorySpider(scrapy.Spider):
name = 'story-spider'
start_urls = ['https://www.storytel.com/in/en/categories/3-Crime?pageNumber=100']
def parse(self, response):
for quote in response.css('div.gridBookTitle'):
item = {
'name': quote.css('a::attr(href)').extract_first()
}
yield item
The original link is https://www.storytel.com/in/en/categories/1-Children. I see that the pageNumber variable is inside script tag, if it helps to find the solution.
Any help would be appreciated. Thanks in advance!!
If you search the XPath like <link rel="next" href=''>
you will found the pagination option. With the help of you can add the pagination code.
here is some example of the pagination page.
next_page = xpath of pagination
if len(next_page) !=0:
next_page_url = main_url.join(next_page
yield scrapy.Request(next_page_url, callback=self.parse)
It will helps you.
I am new to scrapy and writing my first spider make a scrapy spider for website similar to https://blogs.webmd.com/diabetes/default.htm
I want to scrape Headlines and then navigate to each article scrape the text content for each article.
I have tried by using rules and linkextractor but it's not able to navigate to next page and extract. i get the ERROR: Spider error processing https://blogs.webmd.com/diabetes/default.htm> (referer: None)
Below is my code
import scrapy
from scrapy.spiders import Rule
from scrapy.linkextractors import LinkExtractor
class MedicalSpider(scrapy.Spider):
name = 'medical'
allowed_domains = ['https://blogs.webmd.com/diabetes/default.htm']
start_urls = ['https://blogs.webmd.com/diabetes/default.htm']
Rules = (Rule(LinkExtractor(allow=(), restrict_css=('.posts-list-post-content a ::attr(href)')), callback="parse", follow=True),)
def parse(self, response):
headline = response.css('.posts-list-post-content::text').extract()
body = response.css('.posts-list-post-desc::text').extract()
print("%s : %s" % (headline, body))
next_page = response.css('.posts-list-post-content a ::attr(href)').extract()
if next_page:
next_href = next_page[0]
next_page_url = next_href
request = scrapy.Request(url=next_page_url)
yield request
Please guide a newbie in scrapy to get this spider right for multiple articles on each page.
Usually when using scrapy each response is parsed by parse callback. The main parse method is the callback for the initial response obtained for each of the start_urls.
The goal for that parse function should then be to "Identify article links", and issue requests for each of them. Those responses would then be parsed by another callback, say parse_article that would then extract all the contents from that particular article.
You don't even need that LinkExtractor. Consider:
import scrapy
class MedicalSpider(scrapy.Spider):
name = 'medical'
allowed_domains = ['blogs.webmd.com'] # Only the domain, not the URL
start_urls = ['https://blogs.webmd.com/diabetes/default.htm']
def parse(self, response):
article_links = response.css('.posts-list-post-content a ::attr(href)')
for link in article_links:
url = link.get()
if url:
yield response.follow(url=url, callback=self.parse_article)
def parse_article(self, response):
headline = 'some-css-selector-to-get-the-headline-from-the-aticle-page'
# The body is trickier, since it's spread through several tags on this particular site
body = 'loop-over-some-selector-to-get-the-article-text'
yield {
'headline': headline,
'body': body
}
I've not pasted the full code because I believe you still want some excitement learning how to do this, but you can find what I came up with on this gist
Note that the parse_article method is returning dictionaries. These are using Scrapy's items pipelines. You can get a neat json output by running your code using: scrapy runspider headlines/spiders/medical.py -o out.json
My scrapy script seems not to follow links, which ends up not extracting data from each of them (to pass some content as scrapy items).
I am trying to scrape a lot of data from a news website. I managed to copy/write a spider that, as I assumed, should read links from a file (I've generated it with another script), put them in start_urls list and start following these links to extract some data, and then pass it as items, and also -- write each item's data in a separate file (last part is actually for another question).
After running scrapy crawl PNS, script goes through all the links from start_urls but does nothing more -- it follows links read from start_urls list (I'm getting "GET link" message in bash), but seems not to enter them and read some more links to follow and extract data from.
import scrapy
import re
from ProjectName.items import ProjectNameArticle
class ProjectNameSpider(scrapy.Spider):
name = 'PNS'
allowed_domains = ['www.project-domain.com']
start_urls = []
with open('start_urls.txt', 'r') as file:
for line in file:
start_urls.append(line.strip())
def parse(self, response):
for link in response.css('div.news-wrapper_ h3.b-item__title a').xpath('#href').extract():
# extracted links look like this: "/document.html"
link = "https://project-domain.com" + link
yield scrapy.Request(link, callback=self.parse_news)
def parse_news(self, response):
data_dic = ProjectNameArticle()
data_dic['article_date'] = response.css('div.article__date::text').extract_first().strip()
data_dic['article_time'] = response.css('span.article__time::text').extract_first().strip()
data_dic['article_title'] = response.css('h3.article__title::text').extract_first().strip()
news_text = response.css('div.article__text').extract_first()
news_text = re.sub(r'(<script(\s|\S)*?<\/script>)|(<style(\s|\S)*?<\/style>)|(<!--(\s|\S)*?-->)|(<\/?(\s|\S)*?>)', '', news_text).strip()
data_dic['article_text'] = news_text
return data_dic
Expected result:
Script opens start_urls.txt file, reads its lines (every line contains a single link), puts these links to start_urls list,
For each link opened spider extracts deeper links to be followed (that's about 50-200 links for each start_urls link),
Followed links are the main target from which I want to extract specific data: article title, date, time, text.
For now never mind writing each scrapy item to a distinc .txt file.
Actual result:
Running my spider triggers GET for each start_urls link, goes through around 150000, doesn't create a list of deeper links, nor enters them to extract any data.
Dude, I have been coding in Python Scrapy for long time and I hate using start_urls
You can simply use start_requests which is very easy to read, and also very easy to learn for beginners
class ProjectNameSpider(scrapy.Spider):
name = 'PNS'
allowed_domains = ['www.project-domain.com']
def start_requests(self):
with open('start_urls.txt', 'r') as file:
for line in file:
yield Request(line.strip(),
callback=self.my_callback_func)
def my_callback_func(self, response):
for link in response.css('div.news-wrapper_ h3.b-item__title a').xpath('#href').extract():
# extracted links look like this: "/document.html"
link = "https://project-domain.com" + link
yield scrapy.Request(link, callback=self.parse_news)
def parse_news(self, response):
data_dic = ProjectNameArticle()
data_dic['article_date'] = response.css('div.article__date::text').extract_first().strip()
data_dic['article_time'] = response.css('span.article__time::text').extract_first().strip()
data_dic['article_title'] = response.css('h3.article__title::text').extract_first().strip()
news_text = response.css('div.article__text').extract_first()
news_text = re.sub(r'(<script(\s|\S)*?<\/script>)|(<style(\s|\S)*?<\/style>)|(<!--(\s|\S)*?-->)|(<\/?(\s|\S)*?>)', '', news_text).strip()
data_dic['article_text'] = news_text
return data_dic
I also have never used Item class and find it useless too
You can simply have data_dic = {} instead of data_dic = ProjectNameArticle()
I am new to Scrapy and try to use it to practice crawling the website. However, even I followed the codes provided by the tutorial, it does not return the results. It looks like yield scrapy.Request does not work. My codes are as follow:
Import scrapy
from bs4 import BeautifulSoup
from apple.items import AppleItem
class Apple1Spider(scrapy.Spider):
name = 'apple'
allowed_domains = ['appledaily.com']
start_urls =['http://www.appledaily.com.tw/realtimenews/section/new/']
def parse(self, response):
domain = "http://www.appledaily.com.tw"
res = BeautifulSoup(response.body)
for news in res.select('.rtddt'):
yield scrapy.Request(domain + news.select('a')[0]['href'], callback=self.parse_detail)
def parse_detail(self, response):
res = BeautifulSoup(response.body)
appleitem = AppleItem()
appleitem['title'] = res.select('h1')[0].text
appleitem['content'] = res.select('.trans')[0].text
appleitem['time'] = res.select('.gggs time')[0].text
return appleitem
It shows that spider was opened and closed but it returns nothing. The version of Python is 3.6. Can anyone please help? Thanks.
EDIT I
The crawl log can be reached here.
EDIT II
Maybe if I change the codes as below will make the issue clearer:
Import scrapy
from bs4 import BeautifulSoup
class Apple1Spider(scrapy.Spider):
name = 'apple'
allowed_domains = ['appledaily.com']
start_urls = ['http://www.appledaily.com.tw/realtimenews/section/new/']
def parse(self, response):
domain = "http://www.appledaily.com.tw"
res = BeautifulSoup(response.body)
for news in res.select('.rtddt'):
yield scrapy.Request(domain + news.select('a')[0]['href'], callback=self.parse_detail)
def parse_detail(self, response):
res = BeautifulSoup(response.body)
print(res.select('#h1')[0].text)
The codes should print out the url and the title separately but it does not return anything.
Your log states:
2017-07-10 19:12:47 [scrapy.spidermiddlewares.offsite] DEBUG: Filtered offsite request to
'www.appledaily.com.tw': http://www.appledaily.com.tw/realtimenews/article/life/201
70710/1158177/oBike%E7%A6%81%E5%81%9C%E6%A9%9F%E8%BB%8A%E6%A0%BC%E3%80%80%E6%96%B0%E5%8C%
97%E7%81%AB%E9%80%9F%E5%86%8D%E5%85%AC%E5%91%8A6%E5%8D%80%E7%A6%81%E5%81%9C>
Your spider is set to:
allowed_domains = ['appledaily.com']
So it should probably be:
allowed_domains = ['appledaily.com.tw']
It seems like the content you are interested in your parse method (i.e. list items with class rtddt) is generated dynamically -- it can be inspected for example using Chrome, but is not present in HTML source (what Scrapy obtains as a response).
You will have to use something to render the page for Scrapy first. I would recommend Splash together with scrapy-splash package.
I'm trying to get started crawling and scraping a website to disk but having trouble getting the callback function working as I would like.
The code below will visit the start_url and find all the "a" tags on the site. For each 1 of them it will make a callback which is to save the text response to disk and use the crawerItem to store some metadata about the page.
I was hoping someone could help me figure out how to pass
a unique id to each callback so it can be used as the filename when saving the file
Pass the url of the originating page so it can be added to the metadata via the Items
Follow the links on the child pages to go another level deeper into the site
Below is my code thus far
import scrapy
from scrapy.spiders import Rule
from scrapy.linkextractors import LinkExtractor
from mycrawler.items import crawlerItem
class CrawlSpider(scrapy.Spider):
name = "librarycrawler"
allowed_domains = ["example.com"]
start_urls = [
"http://www.example.com"
]
rules = (
Rule(LinkExtractor(),callback='scrape_page', follow=True)
)
def scrape_page(self,response):
page_soup = BeautifulSoup(response.body,"html.parser")
ScrapedPageTitle = page_soup.title.get_text()
item = LibrarycrawlerItem()
item['title'] =ScrapedPageTitle
item['file_urls'] = response.url
yield item
In Settings.py
ITEM_PIPELINES = [
'librarycrawler.files.FilesPipeline',
]
FILES_STORE = 'C:\Documents\Spider\crawler\ExtractedText'
In items.py
import scrapy
class LibrarycrawlerItem(scrapy.Item):
# define the fields for your item here like:
# name = scrapy.Field()
title = scrapy.Field()
Files = scrapy.Field()
I'm not 100% sure but I think you can't rename the scrapy image files however you want, scrapy does that.
What you want to do looks like a job for CrawlSpider instead of Spider.
CrawlSpider by itself follows every link it finds in every page recursively and you can set rules on what pages you want to scrap. Here are the docs.
If you are stubborn enough to keep Spider you can use the meta tag on requests to pass the items and save links in them.
for link in soup.find_all("a"):
item=crawlerItem()
item['url'] = response.urljoin(link.get('href'))
request=scrapy.Request(url,callback=self.scrape_page)
request.meta['item']=item
yield request
To get the item just go look for it in the response:
def scrape_page(self, response):
item=response.meta['item']
In this specific example the item passed item['url'] is obsolete as you can get the current url with response.url
Also,
It's a bad idea to use Beautiful soup in scrapy as it just slows you down, the scrapy library is really well developed to the extent that you don't need anything else to extract data!