How can I find this element with this HTML on Selenium Python?

Question

I'm having a hard time to find a way to select the link below with Selenium by it's name. I need to go by it's name as I will need to identify and select more of it. The identification for this case is: POOL22LATECH

The xpath and CSS selector for this is related to the position on the website and not the name.

Can someone please help?

As an observation, I can't use I can't use the 'tt4', I need to identify by it's name...which, in this case is: POOL22LATECH

Thanks in advance.

HTML:

<a id="tt4" href="javascript:openViewReviewPool('20422', 'Submit for approval', 'fetchPoolAttributes', '1H');">POOL22LATECH<br>20422</a>

XPATH:

//*[@id="tt4"]

CSS Selector:

#poolWithID_3 > td:nth-child(2) > a:nth-child(1)

no...it says: Message: Given xpath expression "//a[contains(.,POOL22LATECH]" is invalid: SyntaxError: Document.evaluate: The expression is not a legal expression — Ana Carolina Araujo, Mar 17 '22 at 20:50
You are missing `'` signs here. Please copy-paste exactly what I gave you — Prophet, Mar 17 '22 at 20:52
Can you share a link to that page or all its HTML, not only the single element HTML? — Prophet, Mar 17 '22 at 21:36

score 0 · Answer 1 · answered Mar 17 '22 at 20:25

0

Have you tried XPath like this ?

"//a[contains(.,'POOL22LATECH')]"

answered Mar 17 '22 at 20:25

Prophet

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score 0 · Answer 2 · answered Mar 17 '22 at 21:23

To locate the element with text POOL22LATECH you can use either of the following locator strategies:

Using partial_link_text:

element = driver.find_element(By.PARTIAL_LINK_TEXT, "POOL22LATECH")

Using css_selector:

element = driver.find_element(By.CSS_SELECTOR, "a[href*='openViewReviewPool'][href*='fetchPoolAttributes']")

Using xpath:

element = driver.find_element(By.XPATH, "//a[contains(@href, 'openViewReviewPool') and contains(., 'POOL22LATECH')]")

To extract the text 2 ideally you need to induce WebDriverWait for the visibility_of_element_located() and you can use either of the following locator strategies:

Using PARTIAL_LINK_TEXT:

element = WebDriverWait(driver, 20).until(EC.visibility_of_element_located((By.PARTIAL_LINK_TEXT, "POOL22LATECH")))

Using CSS_SELECTOR:

element = WebDriverWait(driver, 20).until(EC.visibility_of_element_located((By.CSS_SELECTOR, "a[href*='openViewReviewPool'][href*='fetchPoolAttributes']")))

Using XPATH:

element = WebDriverWait(driver, 20).until(EC.visibility_of_element_located((By.XPATH, "//a[starts-with(@class, 'contactAction') and starts-with(@aria-label, 'Email')][@href]//span[@class]/span")))

Note : You have to add the following imports :

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

score 0 · Accepted Answer · answered Mar 18 '22 at 14:46

0

Thank you all!

I got it with: driver.find_element_by_xpath (f"//*contains(text(),'POOL22LATECH')]")

answered Mar 18 '22 at 14:46

Ana Carolina Araujo

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How can I find this element with this HTML on Selenium Python?

3 Answers3