Click in Selenium + Python

Question

I have a problem with click in Selenium, it doesn't click on the button. This is my code:

from selenium import webdriver
import time
import click
from selenium.webdriver.support.select import Select
from selenium.webdriver.common.keys import Keys


driver = webdriver.Chrome()

driver.get("https://help.instagram.com/contact/723586364339719/")

submit_button = driver.find_element_by_xpath('//*[@id="u_0_8"]')
submit_button.click()

This is the HTML code:

<button value="1" class="_42ft _4jy0 _4jy4 _4jy1 selected _51sy" type="submit" id="u_0_8">Enviar</button>

Ratmir Asanov · Accepted Answer · 2020-03-09 06:29:08Z

1

You can do it with using explicit wait for the button:

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


WebDriverWait(driver, 10).until(EC.element_to_be_clickable((By.XPATH, "//*[@id='u_0_8']"))).click()

I hope it helps you!

edited Mar 9, 2020 at 6:29

answered Mar 8, 2020 at 18:01

Ratmir Asanov

6,4795 gold badges30 silver badges44 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

ParthS007 · Accepted Answer · 2020-03-08 14:00:54Z

0

Instead of this:

submit_button = driver.find_element_by_xpath('//*[@id="u_0_8"]')
submit_button.click()

Try this:

driver.find_element_by_id("id='u_0_8']").click()

answered Mar 8, 2020 at 14:00

ParthS007

2,7191 gold badge26 silver badges39 bronze badges

4 Comments

quique99 Over a year ago

It doesnt work line 242, in check_response raise exception_class(message, screen, stacktrace) selenium.common.exceptions.NoSuchElementException: Message: no such element: Unable to locate element: {"method":"css selector","selector":"[id="id='u_0_8']"]"} (Session info: chrome=80.0.3987.132)

ParthS007 Over a year ago

Try print(driver.page_source), and check the html actually contains the element.

quique99 Over a year ago

i think it actually contains it, no? view-source:help.instagram.com/contact/723586364339719

ParthS007 Over a year ago

You have to print it and check if it contains the same id or not.

undetected Selenium · Accepted Answer · 2020-03-08 21:25:55Z

The desired element is a dynamic element, so 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:

driver.get('https://help.instagram.com/contact/723586364339719/')
WebDriverWait(driver, 10).until(EC.element_to_be_clickable((By.CSS_SELECTOR, "button[type='submit']"))).click()

Using XPATH:

driver.get('https://help.instagram.com/contact/723586364339719/')
WebDriverWait(driver, 10).until(EC.element_to_be_clickable((By.XPATH, "//*[@value='1' and normalize-space()='Send']"))).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

Browser Snapshot:

Collectives™ on Stack Overflow

Click in Selenium + Python

3 Answers 3

Comments

4 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

4 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related