Python 如何从scrapy调用输出文件名

我如何从代码调用输出文件名，即我想在

spider\u closed

函数中使用终端中输入的文件名

scrapy crawl test -o test123.csv

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您可以在spider中使用

self.settings.attributes[“FEED_URI”].value

来获取输出文件名

@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)