CODEWITHSUNDEEP
javaseleniumselenium-webdrivercss-selectors
Sandeep Dharembra
Sandeep Dharembra

Reputation: 312

CSS Locator with contains() InvalidSelectorException using Selenium WebDriver

I am learning Selenium Webdriver and trying to write a simple test script.

The intent is to get the About Google link on Gmail page so as to practice CSS locators.

Here is the code:

public class GoogleSearch {
    
    public static void main(String[] args) {            
        WebDriver driver = new FirefoxDriver();         
        driver.get("https://www.gmail.com");
        
        WebElement aboutGoogle = driver.findElement(By.cssSelector("a:contains('About Google')"));          
    
        driver.close();
        driver.quit();          
    }    
}

I get the below-mentioned exception:

Exception in thread "main" org.openqa.selenium.InvalidSelectorException: The given selector a:contains('About Google') is either invalid or does not result in a WebElement. The following error occurred:
InvalidSelectorError: An invalid or illegal selector was specified
Command duration or timeout: 356 milliseconds
For documentation on this error, please visit: http://seleniumhq.org/exceptions/invalid_selector_exception.html
Build info: version: '2.45.0', revision: '32a636c', time: '2015-03-05 22:01:35'
System info: host: 'XXXXX', ip: '127.0.1.1', os.name: 'Linux', os.arch: 'amd64', os.version: '3.13.0-49-generic', java.version: '1.7.0_79'
*** Element info: {Using=css selector, value=a:contains('Need')}
Session ID: 0f1869f8-c59a-4f61-b1c7-b34ada42573f
Driver info: org.openqa.selenium.firefox.FirefoxDriver

I had checked and was able to find the element in Selenium IDE using the same locator.

I read somewhere that the findElement() method is returning a DOM node and the code is expecting a WebElement object.

If this is the case then, is there a workaround/casting?

Any suggestions?

Upvotes: 10

Views: 52825

Answers (3)

catch32
catch32

Reputation: 18572

The main problem is at this line:

driver.findElement(By.cssSelector("a:contains('About Google')"));

css doesn't maintain contains() for Selenium WD - See here.

For using contains() you have to use Xpath.

With Xpath your locator will be:

//a[contains(text(), 'About Google')]

and for driver it will be as:

driver.findElement(By.xpath("//a[contains(text(), 'About Google')]"));

For finding links with Selenium you can use:

driver.findElement(By.linkText("your link name here"));

It is limitation of CSS selectors compare to Xpath:

  • you can't take parent element with CSS selectors (Xpath has XPath axes)
  • you can't use contains (it is only XPath privilege).

BTW
For processing Xpath locators from page you able to use extension for Firefox browser:

Upvotes: 32

murali selenium
murali selenium

Reputation: 3927

CssSelector does not work in scripting but it works in selenium IDE.

It's also not good to work on sites like gmail.

Upvotes: 6

Anirudh
Anirudh

Reputation: 2326

Well as the Exception is clearly stating the problem here is that your Css Selector is not valid. 'You are trying to get the About Google anchor tag based on it's text which is not a valid css selector'. It's more of a jQuery selector.

You could use the selector based on the value of href attribute as shown below and it will work fine.

 #footer-list a[href*='about']

and use it like 

WebElement aboutGoogle = driver.findElement(By.cssSelector("#footer-list a[href*='about']"));

Upvotes: 0

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