Python 如何在scrapy中访问爬行器中的命令行参数?
我想在Python 如何在scrapy中访问爬行器中的命令行参数?,python,scrapy,Python,Scrapy,我想在scrapy crawl…命令行中传递一个参数,以便在扩展中的规则定义中使用,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
scrapy crawl…
命令行中传递一个参数,以便在扩展中的规则定义中使用,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中的allow属性。
我通过谷歌搜索发现,我可以在spider的
\uuuuu init\uuuu
方法中获取参数值,但是如何在命令行中获取要在规则定义中使用的参数呢?您可以在\uuuu init\uuu
方法中构建spider的规则属性,类似于:
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)
然后在命令行上传递allow
属性,如下所示:
scrapy crawl example.com -a allow="item\.php"