使用scrapy在網址內進行抓取

Question

我正在嘗試使用scrapy抓取craigslist，並且已經成功獲取了url，但是現在我想從url的頁面內提取數據。 以下是代碼：

from scrapy.spider import BaseSpider
from scrapy.selector import HtmlXPathSelector
from craigslist.items import CraigslistItem

class craigslist_spider(BaseSpider):
    name = "craigslist_unique"
    allowed_domains = ["craiglist.org"]
    start_urls = [
        "http://sfbay.craigslist.org/search/sof?zoomToPosting=&query=&srchType=A&addFour=part-time",
        "http://newyork.craigslist.org/search/sof?zoomToPosting=&query=&srchType=A&addThree=internship",
    "http://seattle.craigslist.org/search/sof?zoomToPosting=&query=&srchType=A&addFour=part-time"
    ]


def parse(self, response):
   hxs = HtmlXPathSelector(response)
   sites = hxs.select("//span[@class='pl']")
   items = []
   for site in sites:
       item = CraigslistItem()
       item['title'] = site.select('a/text()').extract()
       item['link'] = site.select('a/@href').extract()
   #item['desc'] = site.select('text()').extract()
       items.append(item)
   hxs = HtmlXPathSelector(response)
   #print title, link        
   return items

我是新手，無法弄清楚如何實際訪問URL（href）並在該URL的頁面中獲取數據並對所有URL執行此操作。

Answer 1

start_urls響應在parse方法中一一收到

如果您只想從該start_urls響應中獲取信息，則您的代碼幾乎可以。 但是您的parse方法應該位於craigslist_spider類中，而不是該類之外。

def parse(self, response):
   hxs = HtmlXPathSelector(response)
   sites = hxs.select("//span[@class='pl']")
   items = []
   for site in sites:
       item = CraigslistItem()
       item['title'] = site.select('a/text()').extract()
       item['link'] = site.select('a/@href').extract()
       items.append(item)
   #print title, link
   return items

如果您想從start_urls中獲取一半信息，並從start_urls響應中存在的anchor中獲取一半信息，該怎么辦？

def parse(self, response):
    hxs = HtmlXPathSelector(response)
    sites = hxs.select("//span[@class='pl']")
    for site in sites:
        item = CraigslistItem()
        item['title'] = site.select('a/text()').extract()
        item['link'] = site.select('a/@href').extract()
        if item['link']:
            if 'http://' not in item['link']:
                item['link'] = urljoin(response.url, item['link'])
            yield Request(item['link'],
                          meta={'item': item},
                          callback=self.anchor_page)


def anchor_page(self, response):
    hxs = HtmlXPathSelector(response)
    old_item = response.request.meta['item'] # Receiving parse Method item that was in Request meta
    # parse some more values
    #place them in old_item
    #e.g
    old_item['bla_bla']=hxs.select("bla bla").extract()
    yield old_item

您只需要在解析方法中yield Request並使用Request meta運送您的old item

然后在old_item中提取old_item ，在anchor_page添加新值並簡單地產生它。

Answer 2

您的xpath中有一個問題-它們應該是相對的。 這是代碼：

from scrapy.item import Item, Field
from scrapy.spider import BaseSpider
from scrapy.selector import HtmlXPathSelector


class CraigslistItem(Item):
    title = Field()
    link = Field()


class CraigslistSpider(BaseSpider):
    name = "craigslist_unique"
    allowed_domains = ["craiglist.org"]
    start_urls = [
        "http://sfbay.craigslist.org/search/sof?zoomToPosting=&query=&srchType=A&addFour=part-time",
        "http://newyork.craigslist.org/search/sof?zoomToPosting=&query=&srchType=A&addThree=internship",
        "http://seattle.craigslist.org/search/sof?zoomToPosting=&query=&srchType=A&addFour=part-time"
    ]

    def parse(self, response):
        hxs = HtmlXPathSelector(response)
        sites = hxs.select("//span[@class='pl']")
        items = []
        for site in sites:
            item = CraigslistItem()
            item['title'] = site.select('.//a/text()').extract()[0]
            item['link'] = site.select('.//a/@href').extract()[0]
            items.append(item)
        return items

如果通過以下方式運行：

scrapy runspider spider.py -o output.json

您將在output.json中看到：

{"link": "/sby/sof/3824966457.html", "title": "HR Admin/Tech Recruiter"}
{"link": "/eby/sof/3824932209.html", "title": "Entry Level Web Developer"}
{"link": "/sfc/sof/3824500262.html", "title": "Sr. Ruby on Rails Contractor @ Funded Startup"}
...

希望能有所幫助。