如何在scrapy函数中删除项目名称？

Question

当运行编码时，它给我的项目价格，但项目名称是相同的。这意味着它先给transcription_price: 245，然后给transcription_price: 240。应该有caption_price和transcription_price。为什么以及如何解决这个问题？

import scrapy
from .. items import FetchingItem
import re

class SiteFetching(scrapy.Spider):
name = 'Site'
start_urls = ['https://www.rev.com/freelancers/transcription',
          'https://www.rev.com/freelancers/captions']

def parse(self, response):
    items = FetchingItem()
    Transcription_price = response.css('#middle-benefit .mt1::text').extract()

    items['Transcription_price'] = Transcription_price

    def next_parse(self, response):
        other_items = FetchingItem()
        Caption_price = response.css('#middle-benefit .mt1::text').extract()

        other_items['Caption_price'] = Caption_price
        yield other_items

    yield items

Answer 1

您的代码永远不会到达方法self.next_parse。默认情况下，Scrapy对self.start_urls中的每个URL调用回调self.parse。您可以通过重写方法start_requests来使用自定义回调。

下面是你如何做到这一点：

import scrapy
from .. items import FetchingItem
import re

class SiteFetching(scrapy.Spider):
    name = 'Site'

    def start_requests(self):
        return [
            scrapy.Request('https://www.rev.com/freelancers/transcription', callback=self.parse_transcription),
            scrapy.Request('https://www.rev.com/freelancers/captions', callback=self.parse_caption)
        ]

    def parse_transcription(self, response):
        items = FetchingItem()
        Transcription_price = response.css('#middle-benefit .mt1::text').extract()

        items['Transcription_price'] = Transcription_price
        yield items

    def parse_caption(self, response):
        other_items = FetchingItem()
        Caption_price = response.css('#middle-benefit .mt1::text').extract()

        other_items['Caption_price'] = Caption_price
        yield other_items

有关详细信息，请参阅Spider documentation。