Scrapy 我可以使用start_url来刮取url列表吗?

Scrapy 我可以使用start_url来刮取url列表吗?,scrapy,Scrapy,我有一个url列表,我想从中获取数据。它来自一个我想更新的数据库,但我不确定如何继续 import scrapy import sqlite3 from datetime import datetime, timedelta class A1hrlaterSpider(scrapy.Spider): name = 'onehrlater' allowed_domains = ['donedeal.ie'] timenow = datetime.now() del

我有一个url列表,我想从中获取数据。它来自一个我想更新的数据库,但我不确定如何继续

import scrapy
import sqlite3
from datetime import datetime, timedelta

class A1hrlaterSpider(scrapy.Spider):
    name = 'onehrlater'
    allowed_domains = ['donedeal.ie']
    timenow = datetime.now()
    delta = timedelta(minutes=0)
    delta2 = timedelta(minutes=1)
    past_time = timenow - delta
    past_time2 = timenow - delta2
    conn = sqlite3.connect('ddother.db')
    c = conn.cursor()
    c.execute("SELECT adUrl FROM database WHERE timestamp BETWEEN ? AND ?", (past_time2, past_time))
    all_urls = c.fetchall()
    urllist = [item[0] for item in all_urls]

    print(urllist)

    conn.commit()

    conn.close()
Urllist是我想要刮取的URL列表。但我不确定如何使用start\u URL 如果这确实是正确的方式,请遵循链接。我可以说start_url=urlist,还是这是错误的


任何帮助都将不胜感激。谢谢

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start\u请求
。好的,我会的,谢谢Gallaecio查看
start\u请求
。好的,我会的,谢谢Gallaecio