<yt-formatted-string id="channel-title" class="style-scope ytd-account-item-renderer">bb1</yt-formatted-string>
below one is a dangerous find element by full x path because the index might change
self.driver.find_element_by_xpath('/html/body/ytd-app/ytd-popup-container/iron-dropdown/div/ytd-multi-page-menu-renderer/div[4]/ytd-multi-page-menu-renderer/div[3]/div[1]/ytd-account-section-list-renderer[1]/div/ytd-account-item-section-renderer/div/ytd-account-item-renderer[4]/paper-icon-item/paper-item-body/yt-formatted-string[1]').click()
To click on the element with text as bb1 you can use either of the following Locator Strategies:
Using css_selector:
self.driver.find_element_by_css_selector("yt-formatted-string.style-scope.ytd-account-item-renderer#channel-title").click()
Using xpath:
self.driver.find_element_by_xpath("//yt-formatted-string[#class='style-scope ytd-account-item-renderer' and #id='channel-title'][text()='bb1']").click()
Ideally, to click on the element you need to induce WebDriverWait for the element_to_be_clickable() and you can use either of the following Locator Strategies:
Using CSS_SELECTOR:
WebDriverWait(self.driver, 20).until(EC.element_to_be_clickable((By.CSS_SELECTOR, "yt-formatted-string.style-scope.ytd-account-item-renderer#channel-title"))).click()
Using XPATH:
WebDriverWait(self.driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//yt-formatted-string[#class='style-scope ytd-account-item-renderer' and #id='channel-title'][text()='bb1']"))).click()
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
References
You can find a couple of relevant detailed discussions in:
How to click a link within youtube comment using python selenium
Related
I am trying to find an element by using Selenium with python, by a known value of src attribute. The element is the point picture on the left of the address of a location in Google Maps.
Namely the html for the element I'm trying to select:
<img alt="" jstcache="935" src="//www.gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp.png" class="Liguzb" jsan="7.Liguzb,0.alt,8.src">
How can I select the given element by searching for it by using the link:
www.gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp.png
Thanks.
To locate the element as the value of src attribute is know to you, you can use either of the following Locator Strategies:
Using css_selector:
element = driver.find_element(By.CSS_SELECTOR, "img.Liguzb[src*='gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp']")
Using xpath:
element = driver.find_element(By.XPATH, "//img[#class='Liguzb' and contains(#src, 'gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp')]")
To locate a visible element instead of presence_of_element_located() you need to induce WebDriverWait for the visibility_of_element_located() and you can use either of the following locator strategies:
Using CSS_SELECTOR:
element = WebDriverWait(driver, 20).until(EC.visibility_of_element_located((By.CSS_SELECTOR, "img.Liguzb[src*='gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp']")))
Using XPATH:
element = WebDriverWait(driver, 20).until(EC.visibility_of_element_located((By.XPATH, "//img[#class='Liguzb' and contains(#src, 'gstatic.com/images/icons/material/system_gm/1x/place_gm_blue_24dp')]")))
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
Can't seem to click on a 'close' button to close a pup up window that I have opened before to scrape the data that it displays.
The button has the following html script:
<div class="closebutton" onclick="return hs.close(this)" title="Schließen"></div>
I was trying the following:
driver.find_element_by_xpath("//div[#class='closebutton']").click()
But I get the error:
Message: element not interactable
Anyone have a clue how to make the element interactable?
To click on the element you can use either of the following Locator Strategies:
Using css_selector:
driver.find_element_by_css_selector("button.closebutton[title='Schließen']").click()
Using xpath:
driver.find_element_by_xpath("//div[#class='closebutton' and #title='Schließen']").click()
Ideally, to click on the element you need to induce WebDriverWait for the element_to_be_clickable() and you can use either of the following Locator Strategies:
Using CSS_SELECTOR:
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.CSS_SELECTOR, "button.closebutton[title='Schließen']"))).click()
Using XPATH:
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//div[#class='closebutton' and #title='Schließen']"))).click()
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
I am scraping data from this website . The element is below and geckodriver
<img class="getdata-button" style="float:right;" src="/common/images/btn-get-data.gif" id="get" onclick="document.getElementById('submitMe').click()">
but can't get selenium to click it tried even xpath, id but not luck
is there any fix or work around to get it done?
To click on the element Get Data you can use either of the following Locator Strategies:
Using css_selector:
driver.find_element_by_css_selector("img.getdata-button#get").click()
Using xpath:
driver.find_element_by_xpath("//img[#class='getdata-button' and #id='get']").click()
Ideally, to click on the element you need to induce WebDriverWait for the element_to_be_clickable() and you can use either of the following Locator Strategies:
Using CSS_SELECTOR:
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.CSS_SELECTOR, "img.getdata-button#get"))).click()
Using XPATH:
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//img[#class='getdata-button' and #id='get']"))).click()
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
You should probably try by id
driver.find_element(By.ID, 'get').click()
I am using Python 3 and Selenium(Chromedriver). I want to check for a element with this command.
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "/html/body/div[3]/div/div[6]/div[2]/div[2]/div/form/div[2]/div[5]/div/div/div/div[2]/button"))).click()
The problem is that the XPATH is constantly changing between two paths:
/html/body/div[3]/div/div[6]/div[2]/div[2]/div/form/div[2]/div[5]/div/div/div/div[2]/button
/html/body/div[3]/div/div[3]/div[2]/div[2]/div/form/div[2]/div[5]/div/div/div/div[2]/button
I want to tell Python, that if the element is not found, it should search for the other XPATH.
If you know a method to find the element without the XPATH, i would also be happy with the solution.
It does not work if you search for the elemnt by its containing text, because the language from the sites changes if use a proxy.
This is the "Inspect Element" code of the button:
<button aria-label="Mobilnummer hinzufügen" class="bg-white css-1eajgu7 ex41m6f0 btn-secondary-dark " type="button">Hinzufügen</button>
To click on the element with text as Hinzufügen you can use either of the following Locator Strategies:
Using css_selector:
driver.find_element_by_css_selector("button.bg-white.btn-secondary-dark[aria-label='Mobilnummer hinzufügen']").click()
Using xpath:
driver.find_element_by_xpath("//button[#aria-label='Mobilnummer hinzufügen' and text()='Hinzufügen']").click()
Ideally, to click on the element you need to induce WebDriverWait for the element_to_be_clickable() and you can use either of the following Locator Strategies:
Using CSS_SELECTOR:
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.CSS_SELECTOR, "button.bg-white.btn-secondary-dark[aria-label='Mobilnummer hinzufügen']"))).click()
Using XPATH:
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//button[#aria-label='Mobilnummer hinzufügen' and text()='Hinzufügen']))).click()
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
Try to use this XPath:
//div[contains(#class, 'mex-mobile-phone')]//button[contains(#class, 'btn-secondary-dark')]
The problem:
I'm trying to click this href here:
Fail attempts:
I tried to use these to no avail
driver.find_element_by_link_text('Join').click()
driver.find_element_by_partial_link_text('href').click()
You can use xpath instead of link text.
driver.find_element_by_xpath('//a[contains(text(), "John"]').click()
Or add space in front of John.
driver.find_element_by_link_text(' Join').click()
To click on the element with text as Join you can use either of the following Locator Strategies:
Using partial_link_text:
driver.find_element_by_partial_link_text("Join").click()
Using xpath:
driver.find_element_by_xpath("//a[contains(., 'Join')]").click()
Ideally, to click on the element you need to induce WebDriverWait for the element_to_be_clickable() and you can use either of the following Locator Strategies:
Using PARTIAL_LINK_TEXT:
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.PARTIAL_LINK_TEXT, "Join"))).click()
Using XPATH:
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//a[contains(., 'Join')]"))).click()
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