WebDriver with Java - click to open a link using xpath

Question

I want to open a a link on a webpage. The link appears to be within a unordered list which resides within in a tag. The url to the web page is selftechy dot com. The tabs are home, about, selenium.

I attempted to open the link using driver.findElement(By.linkText("Selenium")); but page seems like lost its styling. I also tried with xpath method, but it doesn't work either. Please explain to me why it doesn't work and how should I modify the code to make it work properly. Thanks for your help.

HTML code fragment:

<body class="custom">
<div id="container">
<div id="page">
<ul class="menu">
<li class="tab tab-home current"><a href="http://selftechy.com">Home</a></li>
<li class="tab tab-1"><a href="http://selftechy.com/about" title="About">About</a></li>
<li class="tab tab-2"><a href="http://selftechy.com/selenium-2" title="Selenium">Selenium</a></li>
</ul>

webdriver code to open the link

import java.util.List;
import java.util.concurrent.TimeUnit;
import org.junit.*;

import org.junit.Before;
import org.junit.After;
import org.openqa.selenium.*;
import org.openqa.selenium.firefox.FirefoxDriver;
import org.openqa.selenium.support.ui.Select;

public class selftechyTestng 
{
    private WebDriver driver;
    private String baseUrl;

    @Before
    public void setUp() throws Exception
    {
        driver = new FirefoxDriver();
        baseUrl = "http://selftechy.com/";
        driver.manage().timeouts().implicitlyWait(10, TimeUnit.SECONDS);
    }
    @Test
    public void searchElements() throws Exception{
        driver.get(baseUrl);

            //use By.linkText method the page lost its styling
            driver.findElement(By.linkText("Selenium"));

        //use xpath method to open the link doesn't work either 
        List<WebElement> elements = driver.findElements(By.xpath("//div[@id=page]/*[3]")).click(); 
        driver.findElement(By.xpath("//div[@id=page]/*[3]")).click(); 
    }

}

You might have a problem with your xpaths. Before trying to debug your webdriver script, try using firepath to confirm that your xpath definitely works at identifying your screen element. — Vince Bowdren
– Vince Bowdren, Commented May 20, 2013 at 14:42

dirkk · Accepted Answer · 2013-05-19 15:34:16Z

7

Why do you search for the div and then the child element - Is there any particular reason? I don't see any advantage and certainly you are then not getting the a element which you actually want to click. In my opinion it is much simpler to use

driver.findElement(By.xpath("//a[@title = 'Selenium']")).click();

Using your approach you have to use

driver.findElement(By.xpath("//div[@id = 'page']/ul/li[3]/a")).click();

answered May 19, 2013 at 15:34

dirkk

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3 Comments

user2061466 Over a year ago

Thanks for the reply. Both of your solutions worked. I see that @ is used for selecting the attribute within the HTML tag.

Krishna Chandra Tiwari Over a year ago

@dirkk: how would you have identified the same if title was not available as an attribute in html.

dirkk Over a year ago

@KrishnaChandraTiwari If you have a new question, please ask an actual new question. It is too difficult to solve it here in the comments, as it totally depends how your HTML looks like - You might not have a title attribute, but you sure have something else to identify the element.

Nora · Accepted Answer · 2013-05-20 19:05:28Z

0

You can also use this xpath:

"//a[text()='Selenium']"

This will find the link with text = Selenium

answered May 20, 2013 at 19:05

Nora

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Comments

Nagaraju Jampani · Accepted Answer · 2013-10-16 12:40:25Z

Below code will open the link in new window and prints the title and url of the newly opened window.

    String defaultwindow = "";
@Test(description="Main Page")
public void UserOnMainPage()
{
        driver.get("http://yoururl.com");
        defaultwindow = driver.getWindowHandle();
        String selectAll = Keys.chord(Keys.SHIFT,Keys.RETURN);
        driver.findElement(By.linkText("linkname")).sendKeys(selectAll);
        printTitleandUrlofNewlyOpenedwindow();
}

private void printTitleandUrlofNewlyOpenedwindow() 
{
        Set<String> windowHandles1 = driver.getWindowHandles();
        int size = windowHandles1.size();
        System.out.println(size); 
        for (String string : windowHandles1) 
        {
           driver.switchTo().window(string);

           if(string.equals(defaultwindow))
           {
               System.out.println("On Main Window");
               Reporter.log("On Main Window");
           }
           else
           {
               String title=driver.getTitle();
               System.out.println(title);
               Reporter.log(title);  
               String recipeUrl = driver.getCurrentUrl();
               System.out.println(recipeUrl);     
               Reporter.log(recipeUrl);

           }
       }
       driver.switchTo().window(defaultwindow);
}

Nagaraju Jampani · Accepted Answer · 2013-10-16 12:44:28Z

0

Below code will open the link in new Tab.

String selectLinkOpeninNewTab = Keys.chord(Keys.CONTROL,Keys.RETURN); driver.findElement(By.linkText("urlLink")).sendKeys(selectLinkOpeninNewTab);

answered Oct 16, 2013 at 12:44

Nagaraju Jampani

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