When I run the code, the website loads up fine but then it won't click on the button- an error appears saying the element is not interacterble. What do I need to do to click the button? I am relatively new to this and would be grateful for any help.
I have already tried finding it by id and tag.
page = driver.get("https://kenpreston.co.uk/author/")
element = driver.find_element_by_id('mk-button-31')
element.click()
SOLVED:
I used driver.find_element_by_link_text and this worked fine.
I have checked the website and noticed that mk-button-31 is an id for a div tag and inside it there is an a tag. Try getting the url from the a tag and do another driver.get instead of clicking on it.
Also the whole div tag is not clickable so that is why you are getting this error.
Use sleep from time library to be sure page fully loaded
from time import sleep
page = driver.get("https://kenpreston.co.uk/author/")
sleep(2)
element = driver.find_element_by_id('mk-button-31')
element.click()
Looks like your element is not clickable you need to replace this id selector with the css and need to wait for the element before click on it.
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
page = driver.get("https://kenpreston.co.uk/author/")
element = WebDriverWait(driver, 10).until(EC.element_to_be_clickable((By.CSS_SELECTOR, "#mk-button-31 span"))
element.click();
Consider adding Explicit Wait to your script as it might be the case the DOM had finished loading and the button you're looking for is still not there.
The classes you're looking for are:
WebDriverWait
expected_conditions
Suggested code change:
#your other imports here
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions
#your other code here
page = driver.get("https://kenpreston.co.uk/author/")
element = WebDriverWait(driver, 10).until(expected_conditions.element_to_be_clickable((By.ID, "mk-button-31")))
element.click()
More information: How to use Selenium to test web applications using AJAX technology
Related
Can not find the element
The code was writen using python with visual studio code
from time import time
import time
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.support.ui import WebDriverWait
driver = webdriver.Chrome()
paginaHit = 'https://hit.com.do/solicitud-de-verificacion/'
driver.get(paginaHit)
driver.maximize_window()
time.sleep(5)
bl = 'SMLU7318830A'
elementoBL = driver.find_element(By.XPATH, '//*[#id="billoflanding"]').send_keys(bl)
# WebDriverWait(driver,2).until(EC.element_to_be_clickable((By.NAME, "bl"))).Click()
The code is OK, but can not find the element in the webpage.
The portion of the page you are trying to access is inside an EMBED tag. It looks similar to an IFRAME so I would start by switching the context to the EMBED tag and then try searching for the element.
driver = webdriver.Chrome()
paginaHit = 'https://hit.com.do/solicitud-de-verificacion/'
driver.get(paginaHit)
driver.maximize_window()
embed = driver.find_element(By.CSS_SELECTOR, "embed")
driver.switch_to.frame(embed)
bl = 'SMLU7318830A'
wait =WebDriverWait(driver, 20)
wait.until(EC.visibility_of_element_located((By.ID, "billoflanding"))).send_keys(bl)
Couple of additional points:
Don't use sleeps... sleeps are a bad practice. Instead use WebDriverWait when you need to wait for something to happen.
If you are using an ID to find an element, use By.ID and not XPath. ID should be preferred, when available. Next should be a CSS selector and then finally, XPATH only when needed, e.g. to locate elements by contained text or to do complicated DOM traversal.
Hi i'm new at selenium and webscraping and i need some help.
i try to scrape one site and i need and i dont know how to get span class.
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
import time
PATCH = "/Users/bobo/Downloads/chromedriver"
driver = webdriver.Chrome(PATCH)
driver.get("https://neonet.pl")
print(driver.title)
search = driver.find_element_by_class_name("inputCss-input__label-263")
search.send_keys(Keys.RETURN)
time.sleep(5)
i try to extract this span
<span class="inputCss-input__label-263">Szukaj produktu</span>
I can see that you are trying to search something in the search bar.
First I recommend you to use the xpath instead of the class name, here is a simple technique to get the xpath of every element on a webpage:
right-click/ inspect element/ select the mouse in a box element on the upper left/ click on the element on the webpage/ it will directly show you the corresponding html/ then right click on the selected html/ copy options and then xpath.
Here is a code example that searches an element on the webpage, I also included the 'Webdriver-wait' option because sometimes the code runs to fast and can't find the next element so this function make the code wait till the element is visible:
from selenium import webdriver
from selenium.webdriver.common.by import By
from time import sleep
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
driver = webdriver.Chrome(executable_path="/Users/bobo/Downloads/chromedriver")
driver.get("https://neonet.pl") #loading page
wait = WebDriverWait(driver, 20) #defining webdriver wait
search_word = 'iphone\n' # \n is going to act as an enter key
wait.until(EC.visibility_of_element_located((By.XPATH, '//*[#id="root"]/main/div[1]/div[4]/div/div/div[2]/div/button[1]'))).click() #clicking on cookies popup
wait.until(EC.visibility_of_element_located((By.XPATH, '//*[#id="root"]/main/header/div[2]/button'))).click() #clicking on search button
wait.until(EC.visibility_of_element_located((By.XPATH, '//*[#id="root"]/aside[2]/section/form/label/input'))).send_keys(search_word) #searching on input button
print('done!')
sleep(10)
Hope this helped you!
wait=WebDriverWait(driver,10)
driver.get('https://neonet.pl')
elem=wait.until(EC.visibility_of_element_located((By.XPATH, "//span[contains(#class,'inputCss-input')]"))).text
print(elem)
To output the value of the search bar you use .text on the Selenium Webelement.
Imports:
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
I'm trying to click on a link on a webpage that has no ID and no individual class. The only thing to lock it down to is the text 'Sessions'.
I have tried:
driver.find_element_by_xpath("//*[contains(text(),'Sessions')]");
driver.find_element_by_xpath("//*[text()='Sessions']");
Both come back with "No such element".
Edit: I have also tried driver.find_element_by_link_text which also didn't work.
I've tried using the full xpath:
/html/body/div/div/div[1]/div/nav/a[3]
To no avail.
That is a link_Text cause it's between anchor tag, use this :
driver.find_element_by_link_text('Sessions').click()
or
A way more good approach is to use ExplicitWaits :
wait = WebDriverWait(driver, 10)
element = wait.until(EC.element_to_be_clickable((By.LINK_TEXT, 'Sessions')))
element.click()
If you want explicit wait you would need to import the below :
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
If the above gives you NoSuchElementException, I would probably suspect this it is in iframe (See the screenshot first tag - I can see body), if it happens to be then in that case you would need to switch to iframe first and continute with this web element.
Code
wait = WebDriverWait(driver, 10)
wait.until(EC.frame_to_be_available_and_switch_to_it((By.XPATH, "iframe xpath here")))
wait.until(EC.element_to_be_clickable((By.PARTIAL_LINK_TEXT, "Sessions"))).click()
Imports :
I am currently trying to automate a data entry process. Every time I refresh the page the ID tags change and make it impossible. I read that I can use CSS selectors or a possible xpath using tags that don't change. It seems that only the ID tags are the problem.
Below is the HTML code for the button. Every time i try and use the CSS selector i got a no such element exception. I think I am doing it wrong. Please help.
<button type="button" class="dark button secondary" data-automation-id=
"btn-footer-save" data-dojo-attach-event="onclick:saveAndStayPressed"
data-qbo-bind="visible:shouldShowSaveAndStayButton" style>Save</button>
Try this:
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
wait = new WebDriverWait(driver, 20)
wait.until(EC.element_to_be_clickable((By.XPATH, "//button[contains(text(),'Save')]"))).click()
The element gets wrapped in a WebDriverWait until it is interactable, in this case clickable
I've created a script in python using selenium to click on a like button available in a webpage. I've used xpath in order to locate that button and I think I've used it correctly. However, the script doesn't seem to find that button and as a results it throws TimeoutException error pointing at the very line containing the xpath.
As it is not possible to hit that like button without logging in, I expect the script to get the relevant html connected to that button so that I understand I could locate it correctly.
I've tried with:
from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support.wait import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
link = "https://www.instagram.com/p/CBi_eIuAwbG/"
with webdriver.Chrome() as driver:
wait = WebDriverWait(driver,10)
driver.get(link)
item = wait.until(EC.visibility_of_element_located((By.XPATH,"//button[./svg[#aria-label='Like']]")))
print(item.get_attribute("innerHTML"))
How can I locate that like button visible as heart symbol using selenium?
To click on Like Button induce WebDriverWait() and wait for visibility_of_element_located() and below xpath.
Then scroll the element into view and click.
from selenium import webdriver
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.by import By
driver.get("https://www.instagram.com/p/CBi_eIuAwbG/")
element=WebDriverWait(driver,10).until(EC.visibility_of_element_located((By.XPATH,"//button[.//*[name()='svg' and #aria-label='Like']]")))
element.location_once_scrolled_into_view
element.click()
You can do it like this
likeSVG = driver.find_element(By.CSS_SELECTOR, 'svg[aria-label="Like"]')
likeBtn = likeSVG.find_element(By.XPATH, './..')
likeBtn.click()
likeBtn is equal to the parent of the likeSVG div as you can use XPATH similar to file navigation commands in a CLI.
Try using the .find_element_by_xpath(xPath) method (Uses full xpath):
likeXPATH = "/html/body/div[1]/section/main/div/div[1]/article/div[2]/section[1]/span[1]/button"
likeElement = driver.find_element_by_xpath(likeXPATH)
likeElement.click()