我的蜘蛛有一些问题。我使用 scrapy 的splash 来获取由JavaScript 生成的“下一页”的链接。从第一页下载信息后，我想从以下页面下载信息，但 LinkExtractor 功能无法正常工作。但看起来 start_request 函数不起作用。这是代码：

class ReutersBusinessSpider(CrawlSpider):
   name = 'reuters_business'
   allowed_domains = ["reuters.com"]
   start_urls = (
       'http://reuters.com/news/archive/businessNews?view=page&page=1',
   )

    def start_requests(self):
        for url in self.start_urls:
            yield scrapy.Request(url, self.parse, meta={
                'splash': {
                    'endpoint': 'render.html',
                    'args': {'wait': 0.5}
                }
            })
    def use_splash(self, request):
        request.meta['splash'] = {
                'endpoint':'render.html',
                'args':{
                    'wait':0.5,
                    }
                }
        return request

    def process_value(value):
        m = re.search(r'(\?view=page&page=[0-9]&pageSize=10)', value)
        if m:
            return urlparse.urljoin('http://reuters.com/news/archive/businessNews',m.group(1))


    rules = (
        Rule(LinkExtractor(restrict_xpaths='//*[@class="pageNext"]',process_value='process_value'),process_request='use_splash', follow=False),
        Rule(LinkExtractor(restrict_xpaths='//h2/*[contains(@href,"article")]',process_value='process_value'),callback='parse_item'),
    )



    def parse_item(self, response):
        l = ItemLoader(item=PajaczekItem(), response=response)

        l.add_xpath('articlesection','//span[@class="article-section"]/text()', MapCompose(unicode.strip), Join())
        l.add_xpath('date','//span[@class="timestamp"]/text()', MapCompose(parse))
        l.add_value('url',response.url)
        l.add_xpath('articleheadline','//h1[@class="article-headline"]/text()', MapCompose(unicode.title))
        l.add_xpath('articlelocation','//span[@class="location"]/text()')
        l.add_xpath('articletext','//span[@id="articleText"]//p//text()', MapCompose(unicode.strip), Join())

        return l.load_item()

Logs:

2016-02-12 08:20:29 [scrapy] INFO: Spider opened 2016-02-12 08:20:29 [scrapy] INFO: Crawled 0 pages (at 0 pages/min), scraped 0 items (at 0 items/min)
2016-02-12 08:20:29 [scrapy] DEBUG: Telnet console listening on 127.0.0.1:6023
2016-02-12 08:20:38 [scrapy] DEBUG: Crawled (200) <POST localhost:8050/render.html>; (referer: None)
2016-02-12 08:20:38 [scrapy] DEBUG: Filtered offsite request to 'localhost': <GET http://localhost:8050/render.html?page=2&pageSize=10&view=page%3E;
2016-02-12 08:20:38 [scrapy] INFO: Closing spider (finished)

哪里有错误？感谢帮助。

快速浏览一下，您没有使用splash 调用start_request 属性...例如，您应该使用SplashRequest。

def start_requests(self):
    for url in self.start_urls:
        yield SplahRequest(url, self.parse, meta={
            'splash': {
                'endpoint': 'render.html',
                'args': {'wait': 0.5}
            }
        })

假设您已经正确设置了 Splash，即在设置中启用了必要的中间位置并指向正确的 /url 也使它们能够正确启动和 HTTP 缓存...不，我还没有运行您的代码应该很好现在去

编辑：顺便说一句...它不是下一页不是 js 生成的

所以......除非有任何其他原因你使用splash，我认为没有理由在文章请求的初始解析中使用它一个简单的for循环，例如......

for next in response.css("a.control-nav-next::attr(href)").extract():
    yield scrapy.Request(response.urljoin(next), callback=self.parse...