Handling pop-up windows in Selenium Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Selenium code snippet?

Consider the following Selenium Python code that handles a simple alert pop-up. What will be printed after execution?

Selenium Python

from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.common.alert import Alert

# Assume driver is already initialized and navigated to page
alert_button = driver.find_element(By.ID, 'alertBtn')
alert_button.click()
alert = Alert(driver)
alert_text = alert.text
alert.accept()
print(alert_text)

A No output, code raises NoAlertPresentException

B Code raises ElementNotInteractableException

C Empty string printed

D The text message shown in the alert pop-up

💡 Hint

Remember that Alert(driver) switches context to the alert pop-up and .text reads its message.

The Alert object allows reading the alert's text before accepting it. The code prints the alert's message.

❓ assertion

intermediate

Which assertion correctly verifies a confirmation pop-up was accepted?

After clicking a button that triggers a confirmation pop-up, you accept it. Which assertion correctly checks that the pop-up was accepted by verifying the page text?

Selenium Python

from selenium.webdriver.common.by import By

# After accepting confirmation pop-up
result_text = driver.find_element(By.ID, 'result').text

A assert 'Error' in result_text

B assert result_text == 'Cancelled'

C assert 'Confirmed' in result_text

D assert result_text == ''

💡 Hint

Accepting confirmation usually changes page text to indicate confirmation.

The page updates to show 'Confirmed' after accepting the pop-up, so the assertion checks for that text.

❓ locator

advanced

Which locator is best to find a button that opens a new browser window?

You want to locate a button that opens a new window. The button has a unique aria-label 'Open new window'. Which locator is best practice?

A driver.find_element(By.ID, 'openWindowBtn')

B driver.find_element(By.CSS_SELECTOR, 'button[aria-label="Open new window"]')

C driver.find_element(By.CLASS_NAME, 'btn-open')

D driver.find_element(By.XPATH, '//button[text()="Open new window"]')

💡 Hint

Use semantic attributes for accessibility and uniqueness.

Using aria-label with CSS selector is reliable and accessible for locating the button.

🔧 Debug

advanced

Why does this Selenium code fail to switch to the new window?

This code clicks a button to open a new window, then tries to switch to it. Why does it fail?

Selenium Python

original_window = driver.current_window_handle
button = driver.find_element(By.ID, 'openWindowBtn')
button.click()
# Attempt to switch
for handle in driver.window_handles:
    if handle != original_window:
        driver.switch_to.window(handle)
        break
print(driver.title)

A The new window is not fully opened before switching; need to wait

B The loop incorrectly switches back to original window

C driver.window_handles is empty after click

D driver.switch_to.window() requires window name, not handle

💡 Hint

New windows may take time to open; immediate switch can fail.

Without waiting, the new window handle may not appear yet, causing switch to fail.

❓ framework

expert

Which test framework feature best handles flaky pop-up alerts in Selenium tests?

Pop-up alerts sometimes appear with delay causing test failures. Which framework feature best handles this to make tests stable?

A Use explicit waits to wait for alert presence before interacting

B Use implicit waits globally to wait for all elements including alerts

C Wrap alert handling code in try-except without waits

D Disable pop-ups in browser settings to avoid alerts

💡 Hint

Waiting specifically for alerts is more reliable than global waits.

Explicit waits for alert presence ensure code interacts only when alert is ready, reducing flakiness.

Practice

1. What is the correct way to switch to a browser alert pop-up in Selenium with Python?

easy

A. driver.switch_to.window('alert')

B. driver.switch_to.alert

C. driver.switch_to.frame('alert')

D. driver.switch_to.default_content()

Solution

1. Step 1: Understand alert handling in Selenium

   Alerts are special pop-ups that require switching to alert context using driver.switch_to.alert.

2. Step 2: Identify the correct method to switch

   Using driver.switch_to.alert accesses the alert object directly, allowing interaction.

3. Final Answer:

   driver.switch_to.alert -> Option B

4. Quick Check:

   Alert pop-up switch = driver.switch_to.alert [OK]

Hint: Use driver.switch_to.alert for alerts, not window or frame [OK]

Common Mistakes:

• Using switch_to.window for alerts
• Trying to switch to alert as a frame
• Not switching before interacting with alert

2. Which of the following is the correct syntax to switch to a new browser window by its handle in Selenium Python?

easy

A. driver.switch_to.alert(window_handle)

B. driver.switch_to.default_content(window_handle)

C. driver.switch_to.frame(window_handle)

D. driver.switch_to.window(window_handle)

Solution

1. Step 1: Recall method to switch windows

   To switch to a new browser window, use driver.switch_to.window() with the window handle.

2. Step 2: Confirm correct syntax

   The method requires the window handle string as an argument, so driver.switch_to.window(window_handle) is correct.

3. Final Answer:

   driver.switch_to.window(window_handle) -> Option D

4. Quick Check:

   Switch window syntax = driver.switch_to.window(handle) [OK]

Hint: Use switch_to.window(handle) to change browser tabs/windows [OK]

Common Mistakes:

• Using switch_to.alert for windows
• Passing window handle to switch_to.frame
• Omitting the window handle argument

3. Given the following code snippet, what will be printed?

driver.get('https://example.com')
main_window = driver.current_window_handle
# Assume clicking button opens a new window
button = driver.find_element('id', 'openWindow')
button.click()
all_windows = driver.window_handles
for handle in all_windows:
    if handle != main_window:
        driver.switch_to.window(handle)
        print(driver.title)
        driver.close()
        break

# Switch back to main window
driver.switch_to.window(main_window)
print(driver.title)

medium

A. Title of new window followed by title of main window

B. Only title of main window printed twice

C. Only title of new window printed

D. No output due to error

Solution

1. Step 1: Understand window switching and closing

   The code clicks a button that opens a new window, then switches to it, prints its title, closes it, and breaks the loop.

2. Step 2: Switch back and print main window title

   After closing the new window, the code switches back to the main window and prints its title.

3. Final Answer:

   Title of new window followed by title of main window -> Option A

4. Quick Check:

   Print new window title, then main window title [OK]

Hint: Switch to new window, print title, close, then switch back [OK]

Common Mistakes:

• Not switching back to main window before printing
• Trying to print after closing main window
• Assuming both titles print simultaneously

4. Identify the error in the following Selenium Python code for handling an alert pop-up:

alert = driver.switch_to.alert()
alert.accept()

medium

A. Incorrect use of parentheses after switch_to.alert

B. Missing import for Alert class

C. alert.accept() should be alert.dismiss()

D. No error, code is correct

Solution

1. Step 1: Check method usage for alert switching

   driver.switch_to.alert is a property, not a method, so it should not have parentheses.

2. Step 2: Correct the syntax

   Remove parentheses: use alert = driver.switch_to.alert to get the alert object.

3. Final Answer:

   Incorrect use of parentheses after switch_to.alert -> Option A

4. Quick Check:

   switch_to.alert is property, no () [OK]

Hint: Don't add () after switch_to.alert, it's a property [OK]

Common Mistakes:

• Adding parentheses after switch_to.alert
• Confusing accept() with dismiss()
• Not switching to alert before accepting

5. You have a test that opens multiple pop-up windows in sequence. Which approach correctly handles switching to each new window, closing it, and returning to the main window without errors?

hard

A. Store main window handle, loop through all handles, switch to each new window, close it, then switch back to main window after loop

B. Switch to each window and close it inside the loop, then switch back to main window after closing all windows

C. Switch to each new window, close it immediately, and switch back to main window inside the loop after each close

D. Close main window first, then switch to new windows one by one

Solution

1. Step 1: Save main window handle before opening pop-ups

   Store the main window handle to return to it later safely.

2. Step 2: Loop through all window handles, switch, close, and return

   For each new window, switch to it, close it, then switch back to main window inside the loop to avoid stale window errors.

3. Step 3: Avoid closing main window before others

   Closing the main window first causes errors; it should be closed last or not at all during pop-up handling.

4. Final Answer:

   Switch to each new window, close it immediately, and switch back to main window inside the loop after each close -> Option C

5. Quick Check:

   Switch-close-switch back per window to avoid errors [OK]

Hint: Switch-close-switch back inside loop for each pop-up [OK]

Common Mistakes:

• Closing main window before others
• Switching back only after closing all pop-ups
• Not storing main window handle before loop