相关疑难解决方法(0)

我的问题是如何做与前一个问题相同的事情,但在Scrapy 0.14中.

在一些网站上使用一个Scrapy蜘蛛

基本上,我有GUI,它采用域,关键字,标签名称等参数,我想创建一个通用的蜘蛛来抓取那些域中的那些关键字.我通过覆盖蜘蛛管理器类或动态创建蜘蛛来阅读使用旧版scrapy的冲突内容.首选哪种方法,如何实施和调用正确的解决方案？提前致谢.

这是我想要通用的代码.它还使用BeautifulSoup.我把它配对了所以希望没有删除任何关键的理解它.

class MySpider(CrawlSpider):

name = 'MySpider'
allowed_domains = ['somedomain.com', 'sub.somedomain.com']
start_urls = ['http://www.somedomain.com']

rules = (
    Rule(SgmlLinkExtractor(allow=('/pages/', ), deny=('', ))),

    Rule(SgmlLinkExtractor(allow=('/2012/03/')), callback='parse_item'),
)

def parse_item(self, response):
    contentTags = []

    soup = BeautifulSoup(response.body)

    contentTags = soup.findAll('p', itemprop="myProp")

    for contentTag in contentTags:
        matchedResult = re.search('Keyword1|Keyword2', contentTag.text)
        if matchedResult:
            print('URL Found: ' + response.url)

    pass

本教程的脚本(如下)包含两个start_urls.

from scrapy.spider import Spider
from scrapy.selector import Selector

from dirbot.items import Website

class DmozSpider(Spider):
    name = "dmoz"
    allowed_domains = ["dmoz.org"]
    start_urls = [
        "http://www.dmoz.org/Computers/Programming/Languages/Python/Books/",
        "http://www.dmoz.org/Computers/Programming/Languages/Python/Resources/",
    ]

    def parse(self, response):
        """
        The lines below is a spider contract. For more info see:
        http://doc.scrapy.org/en/latest/topics/contracts.html
        @url http://www.dmoz.org/Computers/Programming/Languages/Python/Resources/
        @scrapes name
        """
        sel = Selector(response)
        sites = sel.xpath('//ul[@class="directory-url"]/li')
        items = []

        for site in sites:
            item = Website()
            item['name'] = site.xpath('a/text()').extract()
            item['url'] = site.xpath('a/@href').extract()
            item['description'] = site.xpath('text()').re('-\s[^\n]*\\r')
            items.append(item)

        return …