如何从scrapy调用输出文件名

Question

scrapy crawl test -o test123.csv

如何从代码中调用输出文件名，即我想使用在spider_closed函数中在终端中输入的文件名

@classmethod
def from_crawler(cls, crawler, *args, **kwargs):
    spider = super(MySpider, cls).from_crawler(crawler, *args, **kwargs)
    crawler.signals.connect(spider.spider_closed, signal=scrapy.signals.spider_closed)
def spider_closed(self):
    #read test123.csv (whatever the filename is)

Answer 1

您可以在Spider中使用self.settings.attributes["FEED_URI"].value来获取输出文件名。