How to click a link whose href has a certain substring in Selenium?

Question

The following is a bunch of links ONLY one of them has a substring \"long\" as a value for the attribute href

Answer 1

You can do this:

//first get all the <a> elements
List<WebElement> linkList=driver.findElements(By.tagName("a"));

//now traverse over the list and check
for(int i=0 ; i<linkList.size() ; i++)
{
    if(linkList.get(i).getAttribute("href").contains("long"))
    {
        linkList.get(i).click();
        break;
    }
}

in this what we r doing is first we are finding all the <a> tags and storing them in a list.After that we are iterating the list one by one to find <a> tag whose href attribute contains long string. And then we click on that particular <a> tag and comes out of the loop.

Answer 2

use driver.findElement(By.partialLinkText("long")).click();

Answer 3

I need to click the link who's href has substring "long" in it. How can I do this?

With the beauty of CSS selectors.

your statement would be...

driver.findElement(By.cssSelector("a[href*='long']")).click();

This means, in english,

Find me any 'a' elements, that have the href attribute, and that attribute contains 'long'

You can find a useful article about formulating your own selectors for automation effectively, as well as a list of all the other equality operators. contains, starts with, etc... You can find that at: http://ddavison.io/css/2014/02/18/effective-css-selectors.html

Answer 4

With the help of xpath locator also, you can achieve the same.

Your statement would be:

driver.findElement(By.xpath(".//a[contains(@href,'long')]")).click();

And for clicking all the links contains long in the URL, you can use:-

List<WebElement> linksList = driver.findElements(By.xpath(".//a[contains(@href,'long')]"));

for (WebElement webElement : linksList){
         webElement.click();
}