Python Scrapy - 从 mysql 填充 start_urls-数据库问题

Python Scrapy - populate start_urls from mysql(Python Scrapy - 从 mysql 填充 start_urls)

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问题描述

我正在尝试使用 spider.py 从 MYSQL 表中使用 SELECT 填充 start_url.当我运行scrapy runpider spider.py"时，我没有得到任何输出，只是它没有错误地完成了.

I am trying to populate start_url with a SELECT from a MYSQL table using spider.py. When i run "scrapy runspider spider.py" i get no output, just that it finished with no error.

我已经在 python 脚本中测试了 SELECT 查询，并且 start_url 填充了 MYSQL 表中的条目.

I have tested the SELECT query in a python script and start_url get populated with the entrys from the MYSQL table.

spider.py

from scrapy.spider import BaseSpider
from scrapy.selector import Selector
import MySQLdb


class ProductsSpider(BaseSpider):
    name = "Products"
    allowed_domains = ["test.com"]
    start_urls = []

    def parse(self, response):
        print self.start_urls

    def populate_start_urls(self, url):
        conn = MySQLdb.connect(
                user='user',
                passwd='password',
                db='scrapy',
                host='localhost',
                charset="utf8",
                use_unicode=True
                )
        cursor = conn.cursor()
        cursor.execute(
            'SELECT url FROM links;'
            )
    rows = cursor.fetchall()

    for row in rows:
        start_urls.append(row[0])
    conn.close()

推荐答案

更好的方法是覆盖 start_requests 方法.

A better approach is to override the start_requests method.

这可以查询您的数据库，很像populate_start_urls，并返回一个请求对象.

This can query your database, much like populate_start_urls, and return a sequence of Request objects.

您只需要将 populate_start_urls 方法重命名为 start_requests 并修改以下几行:

You would just need to rename your populate_start_urls method to start_requests and modify the following lines:

for row in rows:
    yield self.make_requests_from_url(row[0])

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