将参数传递给回调函数

Question

def parse(self, response):
    for sel in response.xpath('//tbody/tr'):
        item = HeroItem()
        item['hclass'] = response.request.url.split("/")[8].split('-')[-1]
        item['server'] = response.request.url.split('/')[2].split('.')[0]
        item['hardcore'] = len(response.request.url.split("/")[8].split('-')) == 3
        item['seasonal'] = response.request.url.split("/")[6] == 'season'
        item['rank'] = sel.xpath('td[@class="cell-Rank"]/text()').extract()[0].strip()
        item['battle_tag'] = sel.xpath('td[@class="cell-BattleTag"]//a/text()').extract()[1].strip()
        item['grift'] = sel.xpath('td[@class="cell-RiftLevel"]/text()').extract()[0].strip()
        item['time'] = sel.xpath('td[@class="cell-RiftTime"]/text()').extract()[0].strip()
        item['date'] = sel.xpath('td[@class="cell-RiftTime"]/text()').extract()[0].strip()
        url = 'https://' + item['server'] + '.battle.net/' + sel.xpath('td[@class="cell-BattleTag"]//a/@href').extract()[0].strip()

        yield Request(url, callback=self.parse_profile)

def parse_profile(self, response):
    sel = Selector(response)
    item = HeroItem()
    item['weapon'] = sel.xpath('//li[@class="slot-mainHand"]/a[@class="slot-link"]/@href').extract()[0].split('/')[4]
    return item

好吧,我正在主解析方法中抓取整个表格,我从该表中取了几个字段.其中一个字段是一个网址,我想探索它以获得一大堆字段.如何将已创建的ITEM对象传递给回调函数,以便最终项保留所有字段？

正如上面的代码中所示,我能够保存url中的字段(目前的代码)或只保存表中的字段(只是写yield item)但我不能只生成一个包含所有字段的对象一起.

我试过这个,但很明显,它不起作用.

yield Request(url, callback=self.parse_profile(item))

def parse_profile(self, response, item):
    sel = Selector(response)
    item['weapon'] = sel.xpath('//li[@class="slot-mainHand"]/a[@class="slot-link"]/@href').extract()[0].split('/')[4]
    return item

Answer 1

这就是你使用meta关键字的原因.

def parse(self, response):
    for sel in response.xpath('//tbody/tr'):
        item = HeroItem()
        # Item assignment here
        url = 'https://' + item['server'] + '.battle.net/' + sel.xpath('td[@class="cell-BattleTag"]//a/@href').extract()[0].strip()

        yield Request(url, callback=self.parse_profile, meta={'hero_item': item})

def parse_profile(self, response):
    item = response.meta.get('hero_item')
    item['weapon'] = response.xpath('//li[@class="slot-mainHand"]/a[@class="slot-link"]/@href').extract()[0].split('/')[4]
    yield item

还要注意,这样做sel = Selector(response)是浪费资源,与之前的做法不同,所以我改变了它.它自动映射到responseas response.selector,它也有方便的快捷方式response.xpath.

Answer 2

这是将 args 传递给回调函数的更好方法：

def parse(self, response):
    request = scrapy.Request('http://www.example.com/index.html',
                             callback=self.parse_page2,
                             cb_kwargs=dict(main_url=response.url))
    request.cb_kwargs['foo'] = 'bar'  # add more arguments for the callback
    yield request

def parse_page2(self, response, main_url, foo):
    yield dict(
        main_url=main_url,
        other_url=response.url,
        foo=foo,
    )

来源：https : //docs.scrapy.org/en/latest/topics/request-response.html#topics-request-response-ref-request-callback-arguments