Question

Ні！蜘蛛为什么不通过页面？我使用规则......我做错了什么？ Іt仅适用于一页。这是代码：

# -*- encoding: -*-

class JobSpider(CrawlSpider):
    name = 'superjob'
    allowed_domains = ['superjob.ru']
    start_urls = [
        'http://www.superjob.ru/vacancy/search/?t%5B0%5D=4&sbmit=1&period=7'
    ]

    rules = [
        Rule(SgmlLinkExtractor(allow='/vacancy/search/?',
                               restrict_xpaths=(
                                   u'//a[@class="h_border_none"]/<span>следующая</span>')),
             callback='parse',
             follow=True),
    ]

    def parse(self, response):
        hxs = HtmlXPathSelector(response)
        titles = hxs.select(
                '//*[@id="ng-app"]/div[2]/div/div[2]/div/div[1]/div[2]/div/div/h2/a')
        items = []
        for title in titles:
            item = JobItem()
            item['title'] = title.select('//h2/a/text()').extract()
            items.append(item)
        # return items

Answer 1

要解决的5件事：

restrict_xpaths应指向分页块
callback应该被称为parse()
使用LinkExtractor，SgmlLinkExtractor已弃用
使用xpath()代替select() response.xpath()快捷方式
修复内部XPath表达式 - 只需获取text()

修正版：

# -*- coding: utf-8 -*-
from scrapy.linkextractors import LinkExtractor
from scrapy.spiders import CrawlSpider, Rule


class JobSpider(CrawlSpider):
    name = 'superjob'
    allowed_domains = ['superjob.ru']
    start_urls = [
        'http://www.superjob.ru/vacancy/search/?t%5B0%5D=4&sbmit=1&period=7'
    ]

    rules = [
        Rule(LinkExtractor(allow='/vacancy/search/\?', restrict_xpaths=u'//div[@class="Paginator_navnums"]'),
             callback='parse_item',
             follow=True),
    ]

    def parse_item(self, response):
        titles = response.xpath('//*[@id="ng-app"]/div[2]/div/div[2]/div/div[1]/div[2]/div/div/h2/a')
        for title in titles:
            item = JobItem()
            item['title'] = title.xpath('text()').extract()
            yield item

Scrapy不会超过页面

1 个答案: