Scrapy：如何抓取我從Spider獲得的URL？ exceptions.NameError：未定義全局名稱'parse_detail'

Question

我練習scrapy並有一個問題：我想再次抓取從Spider獲得的鏈接，不知道該怎么做

這是我的代碼：如您所見，我抓取的鏈接將保存在參數中：movie_descriptionTW_URL
我寫了yield Request(movie_descriptionTW, parse_detail)將結果發送到def：

def parse_detail(self, response):
    print(response.url)

但是有一個錯誤：exceptions.NameError：未定義全局名稱'parse_detail'
如何解決呢？
請教我！ 謝謝

from scrapy.spider import Spider
from scrapy.selector import Selector
from yahoo.items import YahooItem
from scrapy.http.request import Request   

class MySpider(Spider):   
    name = "yahoogo"
    start_urls = ["https://tw.movies.yahoo.com/chart.html"]  

    def parse(self, response):
        sel = Selector(response)
        sites = sel.xpath("//tr")
        items = []
        for site in sites:
            item = YahooItem()
            ranking_list = site.xpath("td[@class='c1']/span/text()").extract()
            movie_descriptionTW  = site.xpath("(td[@class='c3']/*//a)[position() < last()-1]/text() | td[@class='c3']/a[1]/text() ").extract()
            movie_descriptionTW_URL = site.xpath("(td[@class='c3']/*//a[2]/@href) | td[@class='c3']/a[1]/@href ").extract()   

            # crawl again!
            yield Request(movie_descriptionTW, parse_detail)

            if ranking_list:    
                items.append(item)
        yield items     

    def parse_detail(self, response):
        print(response.url)

Answer 1

使用self.parse_detail來引用類方法，如下所示：

for url in movie_descriptionTW_URL:
    yield Request(url=url, callback=self.parse_detail)