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Selenium wait until user clicks on button

I am trying to make a python program using selenium that opens the browser and does some stuff, waits for the user to click on a button, then selenium takes over again. I know that I am supposed to use an explicit wait but I do not know how to make selenium wait for the user to click a button on the browser before continuing with my code.
Here is a summary of my code so far:
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.by import By
from selenium.webdriver.support.select import Select
from selenium.webdriver.support.ui import WebDriverWait
driver_path = my_driver_path
driver = webdriver.Chrome(executable_path=driver_path, chrome_options=options)
driver.get(my url)
... (Some code that does automation stuff until a certain point)
#What do I do here?
wait = WebDriverWait(driver, 10)
...(After the user clicks I want the program to resume)

Presumably, something on the website changes when the user clicks on the button. Define that change as the wait condition.
For example, if you expect an element to pop up:
def is_visible(by, identifier):
def condition(driver):
elems = driver.find_elements(by, identifier)
ele = elems and elems[0]
exists_and_displayed = ele and ele.is_displayed()
return exists_and_displayed
return condition
wait.until(is_visible(By.ID, 'added_or_unhidden_after_click'))
Or if you expect something to disappear:
def is_not(condition):
#ft.wraps(condition)
def wrapper(driver):
return not condition(driver)
return wrapper
wait.until(is_not(is_visible(By.ID, 'removed_or_hidden_after_click')))
(typed from memory, might have a syntax error or so, but you should get the idea)
(The built-in expected_conditions as EC basically do these same things, but I run into small issues with them here and there the way they're written, so I prefer to roll my own.)

Selenium Explicit Waits for invisibility_of_element() not waiting at all

I am trying to make Selenium wait until a loader div is invisible.
These are my imports:
from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.common.keys import Keys
from selenium.webdriver.support.ui import WebDriverWait, Select
from selenium.webdriver.support.ui import expected_condition as EC
I first click on a button to open a pop up window with table data in it.
action.click()
action.perform()
After clicking I have to wait until I can press a button to export this data in a file. However, the time I have to wait varies wildly. Sometimes 10 seconds sometimes a few minutes. While this section is loading a loader shows up which prevents me from clicking anywhere on the screen.
I am trying to make Selenium wait until this loader is gone. However, for some reason, the script does not wait at all. Not even the maximum time, which is passed into the explicit wait function.
time.sleep(10)
print("Waiting for button")
wait = WebDriverWait(driver, 30) # I am just testing with 30, it will be a larger value
wait.until(EC.invisibility_of_element((By.XPATH, "//div[#class='loader']")))
print("Finished Waiting for button")
driver.find_element_by_xpath("//button[#class='export']").click()
First I am making Selenium wait 10 seconds so that the loader element can actually show up, which already shows up in 1-2 seconds.
After that I use print statement to check how long the script actually wait.
The script does not wait at all. It immediately continues which then causes an error because the button is not clickable yet.

You are passing a By.locator so instead of invisibility_of_element() you must be using invisibility_of_element_located().
First to wait for the visibility and then for the invisibility of the element you need to:
First induce WebDriverWait for the visibility_of_element_located()
Then induce WebDriverWait for the invisibility_of_element as follows:
WebDriverWait(driver, 30).until(EC.visibility_of_element_located((By.XPATH, "//div[#class='loader']")))
WebDriverWait(driver, 30).until(EC.invisibility_of_element_located((By.XPATH, "//div[#class='loader']")))
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'm trying to automate a process on a website. How do I close a pop up window with selenium, when it is not clear when the popup window will appear?

I am trying to close a pop up window with selenium in python, which is not allowing my code to execute further. But I am unable to do that. There is a a pop up window which asks me if i want to sign up but it keeps popping up at inconsistent times. Is there a method to check wether or not the pop-up window is active?
My code so far:
import selenium
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
import time
PATH = "C:\Program Files (x86)\chromedriver.exe";
driver = webdriver.Chrome(PATH);
driver.get("https://www.investing.com/news/")
time.sleep(3)
accept_cookies = driver.find_element_by_xpath('//*[#id="onetrust-accept-btn-handler"]');
accept_cookies.click();

So your problem is that you don't want the signup popup to display.
I was just poking the site's script, I found a really nice way for you to work around it: set user agent to gene. The script to display popup will be disabled if user agent matches some mobile browsers.
import time
from selenium import webdriver
from selenium.webdriver.chrome.options import Options
opts = Options()
opts.add_argument("user-agent=gene")
driver = webdriver.Chrome(options=opts)
# Adds the cookie into current browser context
driver.get("https://www.investing.com/news/")
time.sleep(60) # wait for a minute to see if the popup happens
driver.quit()
Run this code, observe the page, it will not display the signup popup.
Alternative work around:
Use a Chrome profile that already turned off the Signup popup, will also not display the popup
from selenium import webdriver
options = webdriver.ChromeOptions()
options.add_argument("user-data-dir=C:\\Path") #Path to your chrome profile
If you really need to close the popup without these methods, define a check function with try/except. Before doing any actions, call this function to check if there's popup, then close the popup, save the state to some variable (so you don't check it anymore). Since the function has try/except, it will not raise Exception and your code will run.

You can do this with 2 quick methods that I can think of immediately.
1:
You will use this often
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
then when you want to get an element that needs time to load and you want to do actions with it, you implement it with WebDriverWait :
wait = WebDriverWait(driver, 10)
try:
accept_cookies = wait.until(EC.presence_of_element_located((By.XPATH, "'//*
[#id=\"onetrust-accept-btn-handler\"]'")))
catch:
# probably you have a bad locator and the element doesn't exist or it needs more than 10 sec to load.
else:
# logic here is processed after successfully obtaining the pop up
accept_cookies.click()
I would recommend using ActionChains
from selenium.webdriver.common.action_chains import ActionChains
then after you obtain the pop up element make the click like this:
actions = ActionChains(driver)
actions.move_to_element(accept_cookies).click().perform()
This is the equivalent of moving the mouse pointer to the middle of the pop up close btn and clicking it

Wait time between each action

Hello I am new to python and I am trying to create an automation bot (I am very new to python) that logs into instagram and likes a certain number of posts but I am trying to figure out how to add a delay from when it enters the username information and password but I am not sure how to go about that, Also I would appreciate any feedback/recommendations thank you. Here is the code I have so far
from selenium import webdriver
from selenium.webdriver.common.keys import Keys
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
browser = webdriver.Chrome()
def site_login():
browser.get('https://www.instagram.com/')
browser.find_element_by_name("username").send_keys(‘username’)
browser.find_element_by_name("password").send_keys(“password”)
browser.find_element_by_name("Log In").click() #not sure if this works

The time module should help you
import time
time.sleep(seconds) #Enter the time in seconds here

If you want to wait for a specified period then you can use time.sleep(timeInSeconds) which need import time.
import time
time.sleep(number_of_seconds)
But, however I would like to use the explicit wait. Unlike hard timeout, this will wait until the condition is met and continue the script.
# needed the imports
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.support.ui import WebDriverWait
# wait for the element and click (using xpath locator)
WebDriverWait(browser, 10).until(EC.presence_of_element_located((By.XPATH, "xpath_goes_here"))).click()
# wait for the element and enter value (using css locator)
WebDriverWait(browser, 10).until(EC.presence_of_element_located((By.CSS_SELECTOR, "css_locator_goes_here")))).send_keys("enter input")
# store the element and then perform action
loginButton = WebDriverWait(browser, 10).until(EC.presence_of_element_located((By.CSS_SELECTOR, "css_locator_goes_here"))))
loginButton.click()
you can use either CSS or xpath location strategy.
You can also use implicit wait at the driver level by adding below line of code.
browser.implicitly_wait(10)

Instead of using time.sleep(seconds) , I would recommend to use Explicit wait as the previous comment. Since due to environment changes time can be vary. So its better to use Explicit wait. Using time.sleep(seconds) will sleeps the current thread. It's a bad practice.
Thanks

How to create loop in Python Selenium

I am very new so I would need some help. Can you help me create a loop for the following actions. Click on all buttons and refresh the page and then do it 100 more times for example.enter image description here
my code

Based on the minimal sample you provided, you can refactor this pretty easily. Here's how I would fix that:
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC
# wait on buttons
WebDriverWait(driver, 10).until(EC.presence_of_all_elements_located(
(By.XPATH, "//*[#class='btn default check check green markAsChecked']")))
buttons = driver.find_elements_by_xpath("//*[#class='btn default check check green markAsChecked']")
# click buttons in a loop
for button in buttons:
button.click()
I added a WebDriverWait between buttons, as it's better practice to wait on elements before clicking them.