I can't find the button element to click on it
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I am trying to click on the button "View all details" to expand the details on a restaurant from OpenTable but I keep getting a no element exception. 
from selenium import webdriver
driver = webdriver.Chrome(
'/Library/Python/2.7/site-packages/chromedriver')
url = "https://www.opentable.com/chicago-illinois-restaurant-listings"
driver.get(url)
element = driver.find_element_by_xpath(
'//*[@id="search_results"]/div[2]/div[1]/div/div[2]/div[1]/a')
element.click()
driver.find_element_by_css_selector(
'#overview-section > div:nth-child(4) > div.f9f46391 > button').click()
driver.quit()
python selenium web web-scraping
asked Nov 11 at 16:34
Fidel_Willis
217
add a comment |
up vote
0
down vote
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I am trying to click on the button "View all details" to expand the details on a restaurant from OpenTable but I keep getting a no element exception.
from selenium import webdriver
driver = webdriver.Chrome(
'/Library/Python/2.7/site-packages/chromedriver')
url = "https://www.opentable.com/chicago-illinois-restaurant-listings"
driver.get(url)
element = driver.find_element_by_xpath(
'//*[@id="search_results"]/div[2]/div[1]/div/div[2]/div[1]/a')
element.click()
driver.find_element_by_css_selector(
'#overview-section > div:nth-child(4) > div.f9f46391 > button').click()
driver.quit()
python selenium web web-scraping
asked Nov 11 at 16:34
Fidel_Willis
217
You can try doing thisdriver.find_elements_by_class_name("_7f28dfbb _2c55c63f a8e8f9b4 _06bf3736")[1].click();
– ANISH TIWARI
Nov 11 at 18:03
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to click on the button "View all details" to expand the details on a restaurant from OpenTable but I keep getting a no element exception.
from selenium import webdriver
driver = webdriver.Chrome(
'/Library/Python/2.7/site-packages/chromedriver')
url = "https://www.opentable.com/chicago-illinois-restaurant-listings"
driver.get(url)
element = driver.find_element_by_xpath(
'//*[@id="search_results"]/div[2]/div[1]/div/div[2]/div[1]/a')
element.click()
driver.find_element_by_css_selector(
'#overview-section > div:nth-child(4) > div.f9f46391 > button').click()
driver.quit()
python selenium web web-scraping
asked Nov 11 at 16:34
Fidel_Willis
217
I am trying to click on the button "View all details" to expand the details on a restaurant from OpenTable but I keep getting a no element exception.
from selenium import webdriver
driver = webdriver.Chrome(
'/Library/Python/2.7/site-packages/chromedriver')
url = "https://www.opentable.com/chicago-illinois-restaurant-listings"
driver.get(url)
element = driver.find_element_by_xpath(
'//*[@id="search_results"]/div[2]/div[1]/div/div[2]/div[1]/a')
element.click()
driver.find_element_by_css_selector(
'#overview-section > div:nth-child(4) > div.f9f46391 > button').click()
driver.quit()
python selenium web web-scraping
python selenium web web-scraping
asked Nov 11 at 16:34
Fidel_Willis
217
asked Nov 11 at 16:34
Fidel_Willis
217
asked Nov 11 at 16:34
Fidel_Willis
217
asked Nov 11 at 16:34
Fidel_Willis
217
asked Nov 11 at 16:34
Fidel_Willis
217
217
You can try doing thisdriver.find_elements_by_class_name("_7f28dfbb _2c55c63f a8e8f9b4 _06bf3736")[1].click();
– ANISH TIWARI
Nov 11 at 18:03
add a comment |
You can try doing thisdriver.find_elements_by_class_name("_7f28dfbb _2c55c63f a8e8f9b4 _06bf3736")[1].click();
– ANISH TIWARI
Nov 11 at 18:03
You can try doing this
driver.find_elements_by_class_name("_7f28dfbb _2c55c63f a8e8f9b4 _06bf3736")[1].click();– ANISH TIWARI
Nov 11 at 18:03
You can try doing this
driver.find_elements_by_class_name("_7f28dfbb _2c55c63f a8e8f9b4 _06bf3736")[1].click();– ANISH TIWARI
Nov 11 at 18:03
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
Each result link has target='_blank' attribute. That means that if to click the link details page will be opened in new tab. To handle elements on new tab you should switch to it:
driver.get(url)
current = driver.current_window_handle
driver.find_element_by_css_selector('a.rest-row-name').click()
driver.switch_to.window([tab for tab in driver.window_handles if tab != current][0])
Note that you should also wait for button to became clickable:
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait as wait
from selenium.webdriver.support import expected_conditions as EC
wait(driver, 10).until(EC.element_to_be_clickable((By.XPATH, '//button[.="View all details"]'))).click()
answered Nov 11 at 16:53
Andersson
35.8k103066
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Each result link has target='_blank' attribute. That means that if to click the link details page will be opened in new tab. To handle elements on new tab you should switch to it:
driver.get(url)
current = driver.current_window_handle
driver.find_element_by_css_selector('a.rest-row-name').click()
driver.switch_to.window([tab for tab in driver.window_handles if tab != current][0])
Note that you should also wait for button to became clickable:
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait as wait
from selenium.webdriver.support import expected_conditions as EC
wait(driver, 10).until(EC.element_to_be_clickable((By.XPATH, '//button[.="View all details"]'))).click()
answered Nov 11 at 16:53
Andersson
35.8k103066
add a comment |
up vote
1
down vote
Each result link has target='_blank' attribute. That means that if to click the link details page will be opened in new tab. To handle elements on new tab you should switch to it:
driver.get(url)
current = driver.current_window_handle
driver.find_element_by_css_selector('a.rest-row-name').click()
driver.switch_to.window([tab for tab in driver.window_handles if tab != current][0])
Note that you should also wait for button to became clickable:
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait as wait
from selenium.webdriver.support import expected_conditions as EC
wait(driver, 10).until(EC.element_to_be_clickable((By.XPATH, '//button[.="View all details"]'))).click()
answered Nov 11 at 16:53
Andersson
35.8k103066
add a comment |
up vote
1
down vote
up vote
1
down vote
Each result link has target='_blank' attribute. That means that if to click the link details page will be opened in new tab. To handle elements on new tab you should switch to it:
driver.get(url)
current = driver.current_window_handle
driver.find_element_by_css_selector('a.rest-row-name').click()
driver.switch_to.window([tab for tab in driver.window_handles if tab != current][0])
Note that you should also wait for button to became clickable:
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait as wait
from selenium.webdriver.support import expected_conditions as EC
wait(driver, 10).until(EC.element_to_be_clickable((By.XPATH, '//button[.="View all details"]'))).click()
answered Nov 11 at 16:53
Andersson
35.8k103066
Each result link has target='_blank' attribute. That means that if to click the link details page will be opened in new tab. To handle elements on new tab you should switch to it:
driver.get(url)
current = driver.current_window_handle
driver.find_element_by_css_selector('a.rest-row-name').click()
driver.switch_to.window([tab for tab in driver.window_handles if tab != current][0])
Note that you should also wait for button to became clickable:
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait as wait
from selenium.webdriver.support import expected_conditions as EC
wait(driver, 10).until(EC.element_to_be_clickable((By.XPATH, '//button[.="View all details"]'))).click()
answered Nov 11 at 16:53
Andersson
35.8k103066
answered Nov 11 at 16:53
Andersson
35.8k103066
answered Nov 11 at 16:53
Andersson
35.8k103066
answered Nov 11 at 16:53
Andersson
35.8k103066
35.8k103066
add a comment |
add a comment |
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You can try doing this
driver.find_elements_by_class_name("_7f28dfbb _2c55c63f a8e8f9b4 _06bf3736")[1].click();– ANISH TIWARI
Nov 11 at 18:03