Question

我想在scrapy crawl ...命令行中传递一个参数，以便在扩展CrawlSpider的规则定义中使用，如下所示

name = 'example.com'
allowed_domains = ['example.com']
start_urls = ['http://www.example.com']

rules = (
    # Extract links matching 'category.php' (but not matching 'subsection.php')
    # and follow links from them (since no callback means follow=True by default).
    Rule(SgmlLinkExtractor(allow=('category\.php', ), deny=('subsection\.php', ))),

    # Extract links matching 'item.php' and parse them with the spider's method parse_item
    Rule(SgmlLinkExtractor(allow=('item\.php', )), callback='parse_item'),
)

我希望在命令行参数中指定SgmlLinkExtractor中的allow属性。我用google搜索并发现我可以在spider的__init__方法中获取参数值，但是如何在命令行中获取参数以在规则定义中使用？

Answer 1

您可以在rules方法中构建Spider的__init__属性，例如：

class MySpider(CrawlSpider):

    name = 'example.com'
    allowed_domains = ['example.com']
    start_urls = ['http://www.example.com']

    def __init__(self, allow=None, *args, **kwargs):
        self.rules = (
            Rule(SgmlLinkExtractor(allow=(self.allow,),)),
        )
        super(MySpider, self).__init__(*args, **kwargs)

您在命令行上传递allow属性，如下所示：

scrapy crawl example.com -a allow="item\.php"

如何在scrapy中访问爬行蜘蛛中的命令行参数？

1 个答案: