Question

我想在https://lawyers.justia.com/lawyer/robin-d-gross-39828上抓取该个人的网站和Blog的链接。

我到目前为止：

SELECT  
    A - LAG(A, 1, 0) OVER (PARTITION BY PERSON ORDER BY ID),
    B - LAG(B, 1, 0) OVER (PARTITION BY PERSON ORDER BY ID),
    C - LAG(C, 1, 0) OVER (PARTITION BY PERSON ORDER BY ID),
    D - LAG(D, 1, 0) OVER (PARTITION BY PERSON ORDER BY ID),
    E - LAG(E, 1, 0) OVER (PARTITION BY PERSON ORDER BY ID)
    ....
FROM Table

Answer 1

select r.id role_id
     , e.id
     , role_name
     , role_description
     , contact_name
     , e.employee_mail 
     , contact_phone 
  from rcsa.fs_role r 
  left 
  join rcsa.fs_customer_roles_external e
    on e.fs_role_id = r.id 
  left 
  join rcsa.fs_business_service_contacts c
  on e.employee_mail = c.contact_email  
 where role_type_id = 2 

   and fs_customer_id = :account 

 group 
    by r.id
     , employee_mail

select r.id role_id
     , e.id
     , role_name
     , role_description
     , contact_name
     , e.employee_mail
     , contact_phone 
  from rcsa.fs_role r 
  left 
  join rcsa.fs_customer_roles_external e
    on e.fs_role_id = r.id 

   and fs_customer_id =? 

  left 
  join rcsa.fs_business_service_contacts c
    on c.contact_email = e.employee_mail  
 where role_type_id = 2 

 group 
    by role.id
     , employee_mail

输出：

from bs4 import BeautifulSoup
import requests

headers = {
    'User-Agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:73.0) Gecko/20100101 Firefox/73.0'}

r = requests.get(
    "https://lawyers.justia.com/lawyer/robin-d-gross-39828", headers=headers)
soup = BeautifulSoup(r.text, 'html.parser')

for item in soup.findAll("a", {'data-vars-action': ['ProfileWebsite', 'ProfileBlogPost']}):
    print(item.get("href"))

使用BeautifulSoup + Python在href内抓取超链接

1 个答案: