I'm trying to write a small web crawler with Scrapy.
I wrote a crawler that grabs the URLs of certain links on a certain page, and wrote the links to a csv file. I then wrote another crawler that loops on those links, and downloads some information from the pages directed to from these links.
The loop on the links:
cr = csv.reader(open("linksToCrawl.csv","rb"))
start_urls = []
for row in cr:
start_urls.append("http://www.zap.co.il/rate"+''.join(row[0])[1:len(''.join(row[0]))])
If, for example, the URL of the page I'm retrieving information from is:
http://www.zap.co.il/ratemodel.aspx?modelid=835959
then more information can (sometimes) be retrieved from following pages, like:
http://www.zap.co.il/ratemodel.aspx?modelid=835959&pageinfo=2
("&pageinfo=2" was added).
Therefore, my rules are:
rules = (Rule (SgmlLinkExtractor (allow = ("&pageinfo=\d",
), restrict_xpaths=('//a[#class="NumBtn"]',))
, callback="parse_items", follow= True),)
It seemed to be working fine. However, it seems that the crawler is only retrieving information from the pages with the extended URLs (with the "&pageinfo=\d"), and not from the ones without them. How can I fix that?
Thank you!
You can override parse_start_url() method in CrawlSpider:
class MySpider(CrawlSpider):
def parse_items(self, response):
# put your code here
...
parse_start_url = parse_items
Your rule allows urls with "&pageinfo=\d" . In effect only the pages with matching url will be processed. You need to change the allow parameter for the urls without pageinfo to be processed.
Related
everyone, I've been learning scrapy for a month. I need assistance with following problems:
Suppose there are 100-200 urls and I use Rule to extract further links from those urls and I want to limit the request of those links, like maximum 30 requests for each url. Can I do that?
If I'm searching a keyword on all urls, If the word is found on particular url, then I want scrapy to stop searching from that url and move to next one.
I've tried limiting url but it doesn't work at all.
Thanks, i hope everything is clear.
You can use a process_links callback function with your Rule, this will be passed the list of extracted links from each response, and you can trim it down to your limit of 30.
Example (untested):
class MySpider(CrawlSpider):
name = "test"
allowed_domains = ['example.org']
rules = (
Rule(LinkExtractor(), process_links="dummy_process_links"),
)
def dummy_process_links(self, links):
links = links[:30]
return links
If I understand correctly, and you want stop after finding some word in the page of the response, all you need to do is find the word:
def my_parse(self, response):
if b'word' is in response.body:
offset = response.body.find(b'word')
# do something with it
I'm struggling with Scrapy to output only "hits" to a json file. I'm new at this, so if there is just a link I should review, that might help (I've spent a fair amount of time googling around, still struggling) though code correction tips more welcome:).
I'm working off of the scrapy tutorial (https://doc.scrapy.org/en/latest/intro/overview.html) , with the original code outputing a long list including field names and output like "field: output" where both blanks and found items appear. I'd like only to include links that are found, and output them w/o the field name to a file.
For the following code I am trying, if I issue "scrapy crawl quotes2 -o quotes.json > output.json, it works but the quotes.json is always blank (i.e., including if I do "scrapy crawl quotes2 -o quotes.json").
In this case, as an experiment, I only want to return the URL if the string "Jane" is in the URL (e.g., /author/Jane-Austen):
import scrapy
class QuotesSpider(scrapy.Spider):
name = "quotes2"
start_urls = [
'http://quotes.toscrape.com/tag/humor/',
]
def parse(self, response):
for quote in response.css('a'):
for i in quote.css('a[href*=Jane]::attr(href)').extract():
if i is not None:
print(i)
I've tried "yield" and items options, but am not up to speed enough to make them work. My longer term ambition to go to sites without having to understand the html tree (which may in and of itself be the wrong approach) and look for URLs with specific text in the URL string.
Thoughts? Am guessing this is not too hard, but is beyond me.
Well this is happening because you are printing the items, you have to tell Scrapy explicitly to 'yield' them.
But before that i don't see why you are looping through the anchor nodes instead of that you should loop over the quotes using css or XPath selectors, extract all the author links inside that quote and lastly check if that URL contains a specific String (Jane for you case).
for quote in response.css('.quote'):
jane_url = quote.xpath('.//a[contains(#href, "Jane")]').extract_first()
if jane_url is not None:
yield {
'url': jane_url
}
I need a bit help and your guidance on Scrapy.
My Start_Url is :: http://lighting.philips.co.uk/prof/
Have pasted my code below, which is able to get the links / paths till the below url. But not going beyond that. I need to go to each product's page, listed under the path below. In the "productsinfamily" page the specific products are listed (perhaps within a java script). My Crawler is not able to reach those individual product pages.
http://www.lighting.philips.co.uk/prof/led-lamps-and-tubes/led-lamps/corepro-ledbulb/productsinfamily/
Below is the code for the Crawl spider-
import scrapy
from scrapy.spiders import CrawlSpider, Rule
from scrapy.linkextractors import LinkExtractor
class ProductSearchSpider(CrawlSpider):
name = "product_search"
allowed_domains = ["lighting.philips.co.uk"]
start_urls = ['http://lighting.philips.co.uk/prof/']
rules = (Rule(LinkExtractor(allow=
(r'^https?://www.lighting.philips.co.uk/prof/led-lamps-and-tubes/.*', ),),
callback='parse_page', follow=True),)
def parse_page(self, response):
yield{'URL' : response.url}
You are right that the links are defined in javascript.
If you take a look at the html source, on line 3790 you can see a variable named d75products created. This is later used to populate a template and display the products.
The way I'd approach this would be to extract this data from the source and use the json module to load it. Once you have the data, you can do with it whatever you want.
Another way would be to use something (e.g. a browser) to execute the javascript, and then parse the resulting html. I do think that's unnecessary and overcomplicated though.
Am using Scrapy to browse and collect data, but am finding that the spider is crawling lots of unwanted pages. What I'd prefer the spider to do is just begin from a set of defined pages and then parse the content on those pages and then finish. I've tried to implement a rule like the below but it's still crawling a whole series of other pages as well. Any suggestions on how to approach this?
rules = (
Rule(SgmlLinkExtractor(), callback='parse_adlinks', follow=False),
)
Thanks!
Your extractor is extracting every link because it doesn't have any rule arguments set.
If you take a look at the official documentation, you'll notice that scrapy LinkExtractors have lots of parameters that you can set to customize what your linkextractors extract.
Example:
rules = (
# only specific domain links
Rule(LxmlLinkExtractor(allow_domains=['scrapy.org', 'blog.scrapy.org']), <..>),
# only links that match specific regex
Rule(LxmlLinkExtractor(allow='.+?/page\d+.html)', <..>),
# don't crawl speicific file extensions
Rule(LxmlLinkExtractor(deny_extensions=['.pdf','.html'], <..>),
)
You can also set allowed domains for your spider if you don't want it to wonder off somewhere:
class MySpider(scrapy.Spider):
allowed_domains = ['scrapy.org']
# will only crawl pages from this domain ^
I have a Scrapy Crawler that crawls some guides from a forum.
The forum that I'm trying to crawl the data has got a number of pages.
The problem is that I cannot extract the links that I want to because there aren't specific classes or ids to select.
The url structure is like this one: http://www.guides.com/forums/forumdisplay.php?f=108&order=desc&page=1
Obviously I can change the number after desc&page=1 to 2, 3, 4 and so on but I would like to know what is the best choice to do this.
How can I accomplish that?
PS: This is the spider code
http://dpaste.com/hold/794794/
I can't seem to open the forum URL (always redirects me to another website), so here's a best effort suggestion:
If there are links to the other pages on the thread page, you can create a crawler rule to explicitly follow these links. Use a CrawlSpider for that:
class GuideSpider(CrawlSpider):
name = "Guide"
allowed_domains = ['www.guides.com']
start_urls = [
"http://www.guides.com/forums/forumdisplay.php?f=108&order=desc&page=1",
]
rules = [
Rule(SgmlLinkExtractor(allow=("forumdisplay.php.*f=108.*page=",), callback='parse_item', follow=True)),
]
def parse_item(self, response):
# Your code
...
The spider should automatically deduplicate requests, i.e. it won't follow the same URL twice even if two pages link to it. If there are very similar URLs on the page with only one or two query arguments different (say, order=asc), you can specify deny=(...) in the Rule constructor to filter them out.