Python loop for getting list of elements on few pages -

in application have table users,but table can have more 1 page users. want list users in pages selenium webdriver python. need algorithm gonna count users in first page click on second page count there , until pages done, , return entire list. have idea don't know how realize that, below:

def users(driver):     list_users = []     list_users in range(n):        #selector click on second page        num = num + 1        #getting users 1 page        driver.find_elements_by_css_selector(".even .odd")        #click on second page until pages exist        driver.find_elements_by_link_text("%s" % num)            until nosuchelementexception      return list_users 

i know it's not close working algorithm tried explain want

assuming need total number of people , have logged in...

people = driver.find_element_by_xpath('//*               [@id="frmlist_ohrmlistcomponent"]/div[1]/ul/li[1]').text print people 

yields "1-50 of 53" websites index seen here

that tells total amount of people in list

then total use

print people.split()[2] 

which give value of 53

-------------edit-----------------------

to users loop through tags , pages

from math import ceil pages = ceil(float(people.split()[2])/50)  users = [] p in range(1,int(pages)+1):     driver.execute_script('javascript:submitpage(%s)' % p) #change page     in range(1,51):         try:             user = driver.find_element_by_xpath('//*                            [@id="resulttable"]/tbody/tr[%s]/td[2]' % i)                                            #change tag____^             users.append(user.text)         except:             break  print users