Question

尝试了documentation page上显示的示例scrapy用法（名称下的示例：从单个回调中返回多个请求和项目）

我刚刚将域名更改为指向真实网站：

import scrapy

class MySpider(scrapy.Spider):
    name = 'huffingtonpost'
    allowed_domains = ['huffingtonpost.com/']
    start_urls = [
        'http://www.huffingtonpost.com/politics/',
        'http://www.huffingtonpost.com/entertainment/',
        'http://www.huffingtonpost.com/media/',
    ]

    def parse(self, response):
        for h3 in response.xpath('//h3').extract():
            yield {"title": h3}

        for url in response.xpath('//a/@href').extract():
            yield scrapy.Request(url, callback=self.parse)

但在this gist中发布了ValuError。有什么想法吗？

Answer 1

某些提取的链接是相对的（例如，/news/hillary-clinton/）。你应该把它变成绝对的（http://www.huffingtonpost.com/news/hillary-clinton/

import scrapy

class MySpider(scrapy.Spider):
    name = 'huffingtonpost'
    allowed_domains = ['huffingtonpost.com/']
    start_urls = [
        'http://www.huffingtonpost.com/politics/',
        'http://www.huffingtonpost.com/entertainment/',
        'http://www.huffingtonpost.com/media/',
    ]

    def parse(self, response):
        for h3 in response.xpath('//h3').extract():
            yield {"title": h3}

        for url in response.xpath('//a/@href').extract():
            if url.startswith('/'):
                # transform url into absolute
                url = 'http://www.huffingtonpost.com' + url
            if url.startswith('#'):
                # ignore href starts with #
                continue
            yield scrapy.Request(url, callback=self.parse)

官方scrapy示例中的错误？

1 个答案: