I have HTML that looks like the three following sample statements:
<a href="javascript:__doPostBack('ctl00$FormContent$gvResults','Page$10')">...</a>
<a href="javascript:__doPostBack('ctl00$FormContent$gvResults','Page$12')">12</a>
<a href="javascript:__doPostBack('ctl00$FormContent$gvResults','Page$13')">13</a>
(I'd presently be on pg. 11.)
I don't know the Py/Selenium/Splinter syntax for selecting one of the page numbers in a list and clicking on it to go to that page. (Also, I need to be able to identify the element in the argument as, for example, 'Page$10' or 'Page$12', as seen in the __doPostBack notation. Maybe just a 'next page', in so many words, would be fine, but I don't even know how to do that.)
Thank you for any help.
UPDATE II: Here's the code I have to work from:
import time
import win32ui
import win32api
import win32con
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
from ctypes import *
from selenium.webdriver.common.by import By
driver = webdriver.Chrome()
driver.get('http://[site]');
UPDATE III:
Traceback (most recent call last):
File "montpa_05.py", line 47, in <module>
continue_link = driver.find_element_by_link_text('4')
File "C:\Python27\lib\site-packages\selenium\webdriver\remote\webdriver.py", l
ine 246, in find_element_by_link_text
return self.find_element(by=By.LINK_TEXT, value=link_text)
File "C:\Python27\lib\site-packages\selenium\webdriver\remote\webdriver.py", l
ine 680, in find_element
{'using': by, 'value': value})['value']
File "C:\Python27\lib\site-packages\selenium\webdriver\remote\webdriver.py", l
ine 165, in execute
self.error_handler.check_response(response)
File "C:\Python27\lib\site-packages\selenium\webdriver\remote\errorhandler.py"
, line 164, in check_response
raise exception_class(message, screen, stacktrace)
selenium.common.exceptions.NoSuchElementException: Message: u'no such element\n
(Session info: chrome=28.0.1500.95)\n (Driver info: chromedriver=2.2,platform=
Windows NT 6.1 SP1 x86_64)'
The <a>
element is defined as a link. That means that you can select it by link text.
I don't know Python, but the java syntax would be By.linkText(##)
where ##
is the number you want to click on.