Хотите извлечь ссылки членов, приведенных в «представлении»

Question

хочу извлечь ссылки в 'view' на вкладке 'next' страницы, то есть из n страниц

from bs4 import BeautifulSoup

import requests

r = requests.get('https://old.mciindia.org/InformationDesk/IndianMedicalRegister.aspx')
soup = BeautifulSoup(r.text,'lxml')

for links in soup.find('tr',class_='row'):
    for link in links.find('a',id_='lnkDesc'):
        print link['href']

Answer 1

следует начать:

from selenium import webdriver
from bs4 import BeautifulSoup as bs
import time


url = 'https://old.mciindia.org/InformationDesk/IndianMedicalRegister.aspx'

driver = webdriver.Chrome('C:/chromedriver_win32/chromedriver.exe')
driver.get(url)


driver.find_element_by_xpath("//a[contains(text(),'Year of Registration')]").click()

driver.find_elements_by_css_selector("input[type='text']")[-1].send_keys("2015")
driver.find_element_by_css_selector("input[value='Submit']").click()


soup = bs(driver.page_source, 'html.parser')
table = soup.find('table',{'id':'dnn_ctr588_IMRIndex_GV_Search'})
headers = [ header.text.strip() for header in table.find_all('th') ]

next_page = True
while next_page == True:
    soup = bs(driver.page_source, 'html.parser')
    table = soup.find('table',{'id':'dnn_ctr588_IMRIndex_GV_Search'})
    rows = table.find_all('tr')

    for row in rows:
        if len(row.find_all('td')) == 7:
            data = row.find_all('td')

            name = data[4].text.strip()
            root_url = data[6].a['href'].split("'")[1]
            id_url = data[6].a['href'].split("'")[3]

            link = root_url + 'ViewDetails.aspx?ID=' + id_url

            print ('Name: %-50s\t Link: %s' %(name, link))

    time.sleep(5)
    try:
        driver.find_element_by_xpath("//a[contains(text(),'Next')]").click()
    except:
        print ('No more pages')
        next_page=False

driver.close()