How to get url for onclick="window.open('url')
Budget: $10 – $30 USD
I am creating a site scraping script and want to get the url with xpath.
HTML syntax you want to get onclick="window.open('url') "
I want to get this 'url' part.
The driver uses selenium's chromedriver and also uses BeautifulSoup.
HTML syntax you want to get onclick="window.open('url') "
I want to get this 'url' part.
The driver uses selenium's chromedriver and also uses BeautifulSoup.