Scrapy：在 init 内设置的规则被 CrawlSpider 忽略

Question

几天来一直卡在这个问题上，快要发疯了。

我这样称呼我的 scrapy 蜘蛛：

scrapy crawl example -a follow_links="True"

我传入 "follow_links" 标志来确定是否应该抓取整个网站，或者只抓取我在蜘蛛中定义的索引页面。

在蜘蛛的构造函数中检查这个标志以查看应该设置哪个规则：

def __init__(self, *args, **kwargs):

    super(ExampleSpider, self).__init__(*args, **kwargs)

    self.follow_links = kwargs.get('follow_links')
    if self.follow_links == "True":
        self.rules = (
            Rule(LinkExtractor(allow=()), callback="parse_pages", follow=True),
        )
    else:
        self.rules = (
            Rule(LinkExtractor(deny=(r'[a-zA-Z0-9]*')), callback="parse_pages", follow=False),
        )

如果是"True"，则允许所有链接；如果是 "False"，所有链接都被拒绝。

到目前为止一切顺利，但是这些规则被忽略了。让规则得到遵守的唯一方法是在构造函数之外定义它们。这意味着，像这样的东西可以正常工作：

class ExampleSpider(CrawlSpider):

    rules = (
        Rule(LinkExtractor(deny=(r'[a-zA-Z0-9]*')), callback="parse_pages", follow=False),
    )

    def __init__(self, *args, **kwargs):
        ...

所以基本上，在 __init__ 构造函数中定义规则会导致规则被忽略，而在构造函数外部定义规则会按预期工作。

我无法理解这一点。我的代码如下。

import re
import scrapy

from scrapy.linkextractors import LinkExtractor
from scrapy.spiders import CrawlSpider, Rule
from w3lib.html import remove_tags, remove_comments, replace_escape_chars, replace_entities, remove_tags_with_content


class ExampleSpider(CrawlSpider):

    name = "example"
    allowed_domains = ['example.com']
    start_urls = ['http://www.example.com']    
    # if the rule below is uncommented, it works as expected (i.e. follow links and call parse_pages)
    # rules = (
    #     Rule(LinkExtractor(allow=()), callback="parse_pages", follow=True),
    # )

    def __init__(self, *args, **kwargs):

        super(ExampleSpider, self).__init__(*args, **kwargs)

        # single page or follow links
        self.follow_links = kwargs.get('follow_links')
        if self.follow_links == "True":
            # the rule below will always be ignored (why?!)
            self.rules = (
                Rule(LinkExtractor(allow=()), callback="parse_pages", follow=True),
            )
        else:
            # the rule below will always be ignored (why?!)
            self.rules = (
                Rule(LinkExtractor(deny=(r'[a-zA-Z0-9]*')), callback="parse_pages", follow=False),
            )


    def parse_pages(self, response):
        print("In parse_pages")
        print(response.xpath('/html/body').extract())
        return None


    def parse_start_url(self, response):
        print("In parse_start_url")
        print(response.xpath('/html/body').extract())
        return None

感谢您花时间帮助我解决这个问题。

Answer 1

这里的问题是 CrawlSpider 构造函数 (__init__) 也在处理 rules 参数，所以如果你需要分配它们，你必须先做调用默认构造函数。

换句话说，在调用 super(ExampleSpider, self).__init__(*args, **kwargs) 之前做你需要做的一切：

def __init__(self, *args, **kwargs):
    # setting my own rules
    super(ExampleSpider, self).__init__(*args, **kwargs)

Scrapy：在 init 内设置的规则被 CrawlSpider 忽略

Scrapy: Rules set inside init are ignored by CrawlSpider

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Scrapy：在 __init__ 内设置的规则被 CrawlSpider 忽略

Scrapy: Rules set inside __init__ are ignored by CrawlSpider

python

scrapy

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Scrapy：在 init 内设置的规则被 CrawlSpider 忽略

Scrapy: Rules set inside init are ignored by CrawlSpider