Find element by XPath in Selenium Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to find an element by XPath using Selenium.

Selenium Python

element = driver.find_element_by_[1]("//button[@id='submit']")

Drag options to blanks, or click blank then click option'

Axpath

Bid

Ccss_selector

Dname

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'id' or 'name' instead of 'xpath' when the locator is an XPath expression.

Misspelling the method name.

The method find_element_by_xpath is used to locate elements using XPath expressions.

2fill in blank

medium

Complete the code to find an element by XPath that contains specific text.

Selenium Python

element = driver.find_element_by_xpath("//div[contains(text(), '[1]')]")

Drag options to blanks, or click blank then click option'

ACancel

BClick me

CSubmit

DHome

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect text that does not exist in the element.

Forgetting to put the text inside quotes.

The XPath expression uses contains(text(), 'Submit') to find a div element containing the text 'Submit'.

3fill in blank

hard

Fix the error in the code to correctly find an element by XPath.

Selenium Python

element = driver.find_element_by_xpath("//input[@type='[1]']")

Drag options to blanks, or click blank then click option'

Asubmit

Bbutton

Ccheckbox

Dtext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'button' instead of 'submit' for the type attribute.

Using 'text' which selects text input fields, not submit buttons.

The correct XPath attribute value for a submit button input is type='submit'.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps element texts to their XPath using Selenium.

Selenium Python

elements = driver.find_elements_by_xpath("//a")
links = {el.text: el.get_attribute([1]) for el in elements if el.text [2] ''}

Drag options to blanks, or click blank then click option'

A"href"

B!=

C==

D"src"

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'src' attribute which is for images, not links.

Using '==' instead of '!=' which would select empty text elements.

The attribute to get the link URL is 'href'. The condition el.text != '' filters out elements without visible text.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase element texts to their href attribute if href contains 'example'.

Selenium Python

elements = driver.find_elements_by_xpath("//a")
filtered_links = {el.text[1](): el.get_attribute([2]) for el in elements if el.get_attribute([3]) and 'example' in el.get_attribute('href')}

Drag options to blanks, or click blank then click option'

A.upper

B"href"

D.lower

Attempts:

3 left

💡 Hint

Common Mistakes

Using '.lower()' instead of '.upper()' when uppercase is required.

Checking a wrong attribute name in the condition.

The .upper() method converts text to uppercase. The attribute to get is 'href'. The condition checks if 'href' attribute exists before filtering.