I have a problem with creating valid function which will highlight certain defined elements on the webpage. Please note, that I am coding beginner, and the problem may be either simple environment setup issue, or lack of major knowledge of the javascript/selenium features.
I create my script in Eclipse Neon. To set up an environment I've installed node.js and geckodriver to be able to operate on firefox browser. The beginning of my script is:
var webdriver = require('selenium-webdriver'),
By = webdriver.By
var driver = new webdriver.Builder().forBrowser('firefox').build();
I open the webpage by using driver.get();
and then I simply define the element location by using xPath ex.:
var element = driver.findElement(By.xpath("xPath goes here"));
And now the question begins, what should I do to make WebDriver to highlight this specified element with for ex. red border? While browsing the Stack and other similar pages, the only answers I found was something about using JavaScript Executor in Java syntax, or some webdriver functions using
element.style.backgroundColor = 'red'
but i get console error, that style
or some other part of the syntax is not a function. At this point, I am out of solutions how to make this happen, and I slowly doubt, that I will be able to finish this task without the knowledge about html5/java. Maybe anyone has ever encountered such difficulty and will share the clue?
https://jsfiddle.net/osav574j/ <- I've prepared simplyfied version of my script which may give you a clue of how my full code looks like. The highlight part is probably wrong, it's just to show you, how I thought it may be done, but that's pure assumption.
You should try using executeScript()
as below :-
var element = driver.findElement(By.xpath("xPath goes here"));
driver.executeScript("arguments[0].style.backgroundColor = 'red'", element);