python 提取域名、根域名

谁践踏了优雅 2022-10-19 04:20 207阅读 0赞

## 提取url链接根域名 ##

topRootDomain = (
        '.com', '.la', '.cn', '.io', '.co', '.info', '.net', '.org', '.me', '.mobi',
        '.us', '.biz', '.xxx', '.ca', '.co.jp', '.com.cn', '.net.cn',
        '.org.cn', '.mx', '.tv', '.ws', '.ag', '.com.ag', '.net.ag',
        '.org.ag', '.am', '.asia', '.at', '.be', '.com.br', '.net.br',
        '.bz', '.com.bz', '.net.bz', '.cc', '.com.co', '.net.co',
        '.nom.co', '.de', '.es', '.com.es', '.nom.es', '.org.es',
        '.eu', '.fm', '.fr', '.gs', '.in', '.co.in', '.firm.in', '.gen.in',
        '.ind.in', '.net.in', '.org.in', '.it', '.jobs', '.jp', '.ms',
        '.com.mx', '.nl', '.nu', '.co.nz', '.net.nz', '.org.nz',
        '.se', '.tc', '.tk', '.tw', '.com.tw', '.idv.tw', '.org.tw',
        '.hk', '.co.uk', '.me.uk', '.org.uk', '.vg', ".com.hk")
    
    
    def get_domain_root(url):
        '''获取根域名。'''
        domain_root = ""
        try:
            # 若不是 http或https开头，则补上方便正则匹配规则
            if len(url.split("://")) <= 1 and url[0:4] != "http" and url[0:5] != "https":
                url = "http://" + url
    
            reg = r'[^\.]+(' + '|'.join([h.replace('.', r'\.')
                                         for h in topRootDomain]) + ')$'
            pattern = re.compile(reg, re.IGNORECASE)
    
            parts = parse.urlparse(url)
            host = parts.netloc
            m = pattern.search(host)
            res = m.group() if m else host
            domain_root = "" if not res else res
        except Exception as ex:
            print("error_msg: " + str(ex))
        return domain_root

## 提取url链接子域名 ##

def get_domain(url):
        '''解析域名'''
        domain_root = ""
        try:
            if len(url.split("://")) <= 1 and url[0:4] != "http" and url[0:5] != "https":
                url = "http://" + url
            parts = parse.urlparse(url)
            host = parts.netloc
            domain_root = host if host else ''
        except Exception as ex:
            print("error_msg: " + str(ex))
        return domain_root