How to wait a page is loaded in Python Selenium

Question

try:
    next_page_elem = self.browser.find_element_by_xpath("//a[text()='%d']" % pageno)

except noSuchElementException:
    break
print('page ', pageno)
next_page_elem.click()
sleep(10)

I have a page contains information about reports inside a frame. When i use sleep method to wait for next page to load it works, and without it i kept getting the information on page 1. Beside the sleep is there a better to do this in selenium. I have tried out some previous post that is similar but i think the html i have is quite unique please see example below. Selenium Python: how to wait until the page is loaded? Any help I would be much appreciated. Thanks

 <html>
        <div class="z-page-block-right" style="width:60%;">
        <ul class="pageingclass">
        <li/>
        <li>
        <a class="on" href="javascript:void(0)">1</a>
        </li>
        <li>
        <a onclick=" gotoPage('2','2')" href="javascript:void(0)">2</a>
        </li>
        <li>
        <a class=" next " onclick=" gotoPage('2','2')" href="javascript:void(0)"/>
        </li>
        </ul>
        </div>
    </html>

Does the page URL change when you go from page to page - if yes, can you provide these URLs? Thanks. — alecxe, Mar 21 '16 at 02:25
what is the variable which changes when you go from page to page? page title or any other element ? — Rilwan, Mar 21 '16 at 02:32
there is time stamp refreshes with new time stamp
Last refreshed:13:33:50 AEDT Refresh — Ricard Le, Mar 21 '16 at 02:34

alecxe · Accepted Answer · 2016-03-21T02:55:03.280

4

See this element:

<a class="on" href="javascript:void(0)">1</a>

From what I understand this basically means what page are we currently at.

Since we know the page number beforehand, the idea would be to wait until the element with the current page number becomes present:

from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC

wait = WebDriverWait(driver, 10)
try:
    next_page_elem = self.browser.find_element_by_xpath("//a[text()='%d']" % pageno)

except noSuchElementException:
    break
print('page ', pageno)
next_page_elem.click()

# wait for the page to load
wait.until(
    EC.presence_of_element_located((By.XPATH, "//a[@class = 'on' and . = '%d']" % pageno))
)

Where //a[@class = 'on' and . = '%d'] is an XPath expression that would match a element(s) with class="on" and the text equal to the page number that we paste into the expression via string formatting.

edited Mar 21 '16 at 02:55

answered Mar 21 '16 at 02:34

alecxe

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mate you are a legend. You code has worked. could please explain what is the and is doing inside the class. "//a[@class = 'on' and . = '%d']" Thanks again. – Ricard Le Mar 21 '16 at 02:51
@RicardLe sure, updated the answer and included some explanation. Also see if the answer can be marked as accepted to resolve the topic. Thanks! – alecxe Mar 21 '16 at 02:55
Hi, i use fireFox FirePath with the following //a[@class = 'on' = 2] and it has selected the right path. thanks – Ricard Le Mar 21 '16 at 03:03

How to wait a page is loaded in Python Selenium

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