CSS Locator with contains() InvalidSelectorExcepti

问题:

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:

import org.openqa.selenium.By;
import org.openqa.selenium.WebDriver;
import org.openqa.selenium.WebElement;
import org.openqa.selenium.firefox.FirefoxDriver;

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?

回答1:

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

One more is, its not good to work on sites like gmail.. you can learn by using components in http://seleniumtrainer.com/. its good for start up.

thank you

回答2:

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:

• FirePath

• Xpath Checker

回答3:

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']"));