I am working on a class project and trying to get all IMDB movie data (titles, budgets. etc.) up until 2016. I adopted the code from https://github.com/alexwhb/IMDB-spider/blob/master/tutorial/spiders/spider.py.
My thought is: from i in range(1874,2016) (since 1874 is the earliest year shown on http://www.imdb.com/year/), direct the program to the corresponding year's website, and grab the data from that url.
But the problem is, each page for each year only show 50 movies, so after crawling the 50 movies, how can I move on to the next page? And after crawling each year, how can I move on to next year? This is my code for the parsing url part so far, but it is only able to crawls 50 movies for a particular year.
class tutorialSpider(scrapy.Spider):
name = "tutorial"
allowed_domains = ["imdb.com"]
start_urls = ["http://www.imdb.com/search/title?year=2014,2014&title_type=feature&sort=moviemeter,asc"]
def parse(self, response):
for sel in response.xpath("//*[@class='results']/tr/td[3]"):
item = MovieItem()
item['Title'] = sel.xpath('a/text()').extract()[0]
item['MianPageUrl']= "http://imdb.com"+sel.xpath('a/@href').extract()[0]
request = scrapy.Request(item['MianPageUrl'], callback=self.parseMovieDetails)
request.meta['item'] = item
yield request
You can use CrawlSpiders to simplify your task. As you'll see below,
start_requests
dynamically generates the list of URLs whileparse_page
only extracts the movies to crawl. Finding and following the 'Next' link is done by therules
attribute.I agree with @Padraic Cunningham that hard-coding values is not a great idea. I've added spider arguments so that you can call:
scrapy crawl imdb -a start=1950 -a end=1980
(the scraper will default to 1874-2016 if it doesn't get any arguments).I figured out a very dumb way to solve this. I put all the links in the start_urls. Better solution would be very much appreciated!