CODEWITHSUNDEEP
pythonselenium
MQ1217
MQ1217

Reputation: 37

Python Selenium error: NoSuchElementException

I am trying to locate an href containing the substring '.ics', such as in the screenshot, and return the link as a string. screenshot

Below is my code attempt:

from selenium import webdriver
from selenium.webdriver.common.keys import Keys

driver = webdriver.Chrome()
driver.get('http://miamioh.edu/emss/offices/career-services/events/index.html')

element = driver.find_element_by_partial_link_text('.ics')

However, I get this error: 

selenium.common.exceptions.NoSuchElementException: Message: no such element: Unable to locate element: {"method":"partial link text","selector":".ics"}

No doubt I am overlooking something very basic, but I can't figure out what. I have also tried the line

element = driver.findElement(By.cssSelector("a[href*='services.ics']")).click();

instead of the other line beginning with 'element'. However, this gives

AttributeError: 'WebDriver' object has no attribute 'findElement'

Upvotes: 0

Views: 1529

Answers (4)

Andersson
Andersson

Reputation: 52665

The link text is exact text of the link you see on web page while partial link text is just some substring of that link text.

"services.ics" is part of href attribute. If you want to find element by "services.ics" you might use 

driver.find_element_by_xpath('//a[contains(@href, "services.ics")]')

Also you might use title attribute to match required element:

driver.find_element_by_xpath('//a[@title="iCal Feed"]')

Note that 

element = driver.findElement(By.cssSelector("a[href*='services.ics']")).click();

is Java analogue of Python code

from selenium.webdriver.common.by import By 
element = driver.find_element(By.CSS_SELECTOR, "a[href*='services.ics']").click();

Update

Link might be generated dynamically, so you can try to apply ExplicitWait as below to avoid NoSuchElementException:

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

wait(driver, 10).until(EC.element_to_be_clickable((By.XPATH, '//a[@title="iCal Feed"]'))).click()

Update 2

As target link located inside an iframe you should switch to that frame before clicking link:

wait(driver, 10).until(EC.frame_to_be_available_and_switch_to_it("trumba.spud.1.iframe"))
driver.find_element_by_xpath('//a[@title="iCal Feed"]').click()

Upvotes: 4

IamBatman
IamBatman

Reputation: 1015

You need to get the attribute href. First locate that element without looking for the "HREF", then get the attribute.

myHrefVal = element.get_attribute('href')
print(myHrefVal)

As @Andersson has stated, try using the xPath.

driver.find_element_by_xpath('//a[@title="iCal Feed"]')

Good luck! :)

Upvotes: 1

willeM_ Van Onsem
willeM_ Van Onsem

Reputation: 476659

Since you want to search for a <a href="..."> that ends with .ics. We can do this with a CSS selector, like:

a[href$=".ics"]

So we can use the following code:

element = driver.find_element_by_css_selector('a[href$=".ics"]')

Or if you are looking for 'services.ics':

element = driver.find_element_by_css_selector('a[href$="services.ics"]')

Upvotes: 1

TitusLucretius
TitusLucretius

Reputation: 181

In Python the Selenium method is driver.find_element. Also partial link text does not refer to the link, instead it refers to the text in the "a" tag. i.e. 

<a href="link.com">this is the link text</a>

Upvotes: 1

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