python - Scrapy:通过管道从数据库获取Start_Urls

Question

不幸的是，我没有足够的人口来发表评论，所以我必须提出这个新问题，引用https://stackoverflow.com/questions/23105590/how-to-get-the-pipeline-object-in-scrapy-spider

我在数据库中有很多网址。所以我想从我的数据库获取 start_url。到目前为止还不是大问题。好吧，我不希望 mysql 的东西在蜘蛛里面，在管道中我遇到了问题。如果我尝试将管道对象移交给我的蜘蛛，就像所提到的问题一样，我只会收到带有消息的属性错误

'None Type' object has no attribute getUrl

我认为实际的问题是函数spider_opened没有被调用(还插入了一条打印语句，该语句从未在控制台中显示其输出)。有人知道如何在蜘蛛内部获取管道对象吗？

MySpider.py

def __init__(self):
    self.pipe = None

def start_requests(self):
    url = self.pipe.getUrl()
    scrapy.Request(url,callback=self.parse)

管道.py

@classmethod
def from_crawler(cls, crawler):
        pipeline = cls()
        crawler.signals.connect(pipeline.spider_opened, signals.spider_opened)
def spider_opened(self, spider):
    spider.pipe = self

def getUrl(self):
     ...

Answer 1

Scrapy 管道已经具有 open_spider 和 close_spider 的预期方法

取自文档:https://doc.scrapy.org/en/latest/topics/item-pipeline.html#open_spider

open_spider(self, spider)
This method is called when the spider is opened.
Parameters: spider (Spider object) – the spider which was opened

close_spider(self, spider)
This method is called when the spider is closed. Parameters: spider (Spider object) – the spider which was closed

但是您最初的问题没有多大意义，为什么要为您的蜘蛛分配管道引用？这似乎是一个非常糟糕的主意。

你应该做的是打开数据库并读取你的蜘蛛本身的网址。

from scrapy import Spider
class MySpider(Spider):
    name = 'myspider'
    start_urls = []

    @classmethod
    def from_crawler(self, crawler, *args, **kwargs):
        spider = super().from_crawler(crawler, *args, **kwargs)
        spider.start_urls = self.get_urls_from_db()
        return spider

    def get_urls_from_db(self):
        db = # get db cursor here
        urls = # use cursor to pop your urls
        return urls