我需要拉出职业及其对应的专长。 但我的代码只适用于拉动职业。
import requests, bs4
r = requests.get('https://www.doximity.com/sign_ups/9e016f85-d589-4cdf-8240-09c356d4434f/edit?sign_up[user_attributes][firstname]=Jian&sign_up[user_attributes][lastname]=Cui')
soup = bs4.BeautifulSoup(r.text, 'lxml')
spec = soup.find_all('select')
for sub in spec:
print (sub.text)
请给我一些想法。
最佳答案
检查下面的代码，如有任何问题请告诉我:
from selenium import webdriver
from selenium.webdriver.support.ui import Select
import time
driver = webdriver.Chrome()
url = 'https://www.doximity.com/sign_ups/9e016f85-d589-4cdf-8240-09c356d4434f/edit?sign_up[user_attributes][firstname]=Jian&sign_up[user_attributes][lastname]=Cui'
driver.get(url)
spec = driver.find_element_by_id("sign_up_user_attributes_credential_id")
for sub in spec.find_elements_by_xpath('./option | ./optgroup/option'):
if sub.get_attribute('value') != '':
print(sub.text)
selected_spec = Select(driver.find_element_by_id("sign_up_user_attributes_credential_id"))
selected_spec.select_by_visible_text(sub.text)
time.sleep(0.5)
occup = driver.find_element_by_xpath('//select[@id="sign_up_user_attributes_user_professional_detail_attributes_specialty_id"]')
for oc in occup.find_elements_by_xpath('./option'):
if oc.text != '' and oc.get_attribute('value') != '':
print(oc.text)
关于Python 网页抓取 : I have a website with pick list. 以及如何提取这些列表中的文本，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42170243/