How to click on All links in web page in Selenium WebDriver

Question

I have 10 different pages contain different links. How to click on all the links?

Conditions are : i) I don't know how many links are there ii) I want to count and click on each link

Please suggest me Selenium WebDriver script.

score 9 · Accepted Answer · edited May 07 '14 at 19:46

Capture and Navigate all the Links on Webpage

Iterator and advanced for loop can do similar job; However, the inconsistency on page navigation within a loop can be solved using array concept.

private static String[] links = null;
private static int linksCount = 0;

driver.get("www.xyz.com");
List<WebElement> linksize = driver.findElements(By.tagName("a")); 
linksCount = linksize.size();
System.out.println("Total no of links Available: "+linksCount);
links= new String[linksCount];
System.out.println("List of links Available: ");  
// print all the links from webpage 
for(int i=0;i<linksCount;i++)
{
links[i] = linksize.get(i).getAttribute("href");
System.out.println(all_links_webpage.get(i).getAttribute("href"));
} 
// navigate to each Link on the webpage
for(int i=0;i<linksCount;i++)
{
driver.navigate().to(links[i]);
Thread.sleep(3000);
}

1| Capture all links under specific frame|class|id and Navigate one by one

driver.get("www.xyz.com");  
WebElement element = driver.findElement(By.id(Value));
List<WebElement> elements = element.findElements(By.tagName("a"));
int sizeOfAllLinks = elements.size();
System.out.println(sizeOfAllLinks);
for(int i=0; i<sizeOfAllLinks ;i++)
{
System.out.println(elements.get(i).getAttribute("href"));
}   
for (int index=0; index<sizeOfAllLinks; index++ ) {
getElementWithIndex(By.tagName("a"), index).click();
driver.navigate().back();
}

public WebElement getElementWithIndex(By by, int index) {
WebElement element = driver.findElement(By.id(Value));
List<WebElement> elements = element.findElements(By.tagName("a")); 
return elements.get(index);
}

2| Capture all links [Alternate method]

Java

driver.get(baseUrl + "https://www.google.co.in");
List<WebElement> all_links_webpage = driver.findElements(By.tagName("a")); 
System.out.println("Total no of links Available: " + all_links_webpage.size());
int k = all_links_webpage.size();
System.out.println("List of links Available: ");
for(int i=0;i<k;i++)
{
if(all_links_webpage.get(i).getAttribute("href").contains("google"))
{
String link = all_links_webpage.get(i).getAttribute("href");
System.out.println(link);
}   
}

Python

from selenium import webdriver

driver = webdriver.Firefox()
driver.get("https://www.google.co.in/")
list_links = driver.find_elements_by_tag_name('a')

for i in list_links:
        print i.get_attribute('href')

driver.quit()

score 3 · Answer 2 · edited Feb 18 '18 at 15:19

public static void main(String[] args) 
    {
        FirefoxDriver fd=new FirefoxDriver();
        fd.get("http:www.facebook.com");
        List<WebElement> links=fd.findElements(By.tagName("a"));
        System.out.println("no of links:" +links.size());

        for(int i=0;i<links.size();i++)
        {
            if(!(links.get(i).getText().isEmpty()))
            {
            links.get(i).click();
            fd.navigate().back();
            links=fd.findElements(By.tagName("a"));
            }       
        }
   }

This program clicks on a link, navigates back to the page and clicks the second link again.

score 0 · Answer 3 · edited Mar 07 '14 at 19:47

0

Store them in an array then click them:

   ArrayList<WebElement> input_type = (ArrayList<WebElement>)    
  driver.findElements(By.tagName("a"));

 for (WebElement type : input_type)
 {


      type.click();


  }

This will click one by one all links with a tag ,i hope you got the point .Enjoy!

edited Mar 07 '14 at 19:47

ThiefMaster

310,957
84
592
636

answered Feb 20 '14 at 05:19

Himanshu Tewari

313
2
12

ArrayList input_type = (ArrayList) --- This line shows error for me. – Ripon Al Wasim Apr 25 '19 at 05:59

score 0 · Answer 4 · edited Feb 16 '15 at 12:24

import java.util.List;

import org.openqa.selenium.By;

import org.openqa.selenium.WebDriver;

import org.openqa.selenium.WebElement;

import org.openqa.selenium.firefox.FirefoxDriver;

import org.openqa.selenium.firefox.FirefoxProfile;

import org.openqa.selenium.firefox.internal.ProfilesIni;

public class Find_all_Links {

private static String testUrl = "http://www.google.co.in/";

private static WebDriver driver = null;

public static void main(String[] args) {

    ProfilesIni profile = new ProfilesIni();

    FirefoxProfile myProfile = profile.getProfile("AutomationQA");

    driver = new FirefoxDriver(myProfile);

    driver.get(testUrl);

    List<WebElement> oLinksOnPage = driver.findElements(By.tagName("a"));

    System.out.println(oLinksOnPage.size());

    for(int i=0;i<oLinksOnPage.size();i++){

        System.out.println(oLinksOnPage.get(i).getText());
    }


}

}

score 0 · Answer 5 · answered Oct 26 '16 at 06:13

package selenium.tests;

import java.util.List;



import java.util.concurrent.TimeUnit;
import org.openqa.selenium.*;
import org.openqa.selenium.chrome.ChromeDriver;
import org.openqa.selenium.support.ui.ExpectedConditions;
import org.openqa.selenium.support.ui.WebDriverWait;

public class TestAllLinks { 



 public static void main(String[] args) {
        String baseUrl = "http://www.qaautomated.com/";
        System.setProperty("webdriver.chrome.driver", 

         "C:\\Users\\chromedriver_win32\\chromedriver.exe");
        WebDriver driver=new ChromeDriver();
        String notWorkingUrlTitle = "Under Construction: QAAutomated";
        driver.manage().timeouts().implicitlyWait(10, TimeUnit.SECONDS);

        driver.get(baseUrl);
        List<WebElement> linkElements = driver.findElements(By.tagName("a"));
        String[] linkTexts = new String[linkElements.size()];
        int i = 0;

        //extract the link texts of each link element
        for (WebElement elements : linkElements) {
            linkTexts[i] = elements.getText();
            i++;
        }

        //test each link
        for (String t : linkTexts) {
            driver.findElement(By.linkText(t)).click();
            if (driver.getTitle().equals(notWorkingUrlTitle )) {
                System.out.println("\"" + t + "\""
                        + " is not working.");
            } else {
                System.out.println("\"" + t + "\""
                        + " is working.");
            }
            driver.navigate().back();
        }
        driver.quit();
    }
}

http://www.qaautomated.com/2016/10/selenium-test-to-check-links-in-web.html

score 0 · Answer 6 · edited Mar 20 '18 at 03:11

You can use 2 logics to handle

Get the link and hit it in your browser and validate
Get the link and use REST Web service to validate link.

Using REST WS would be the easiest way to validate links.

Below code works fine for me.

public class Testing{

    public static void main(String[] args) {
        try{
            WebDriver driver = new FirefoxDriver();
            driver.manage().timeouts().implicitlyWait(10, TimeUnit.SECONDS);
            driver.manage().window().maximize();
            driver.navigate().to("https://www.amazon.in/");
            List<WebElement> links = driver.findElements(By.tagName("a"));
            System.out.println("Number of links : " + links.size());
            for(WebElement e : links) {
                String linkText = e.getAttribute("href");
                System.out.println("Link -->>" +linkText);
                if(linkText!=null && !linkText.isEmpty()) {
                    HttpPost post = new HttpPost(linkText);
                    HttpClient client = HttpClientBuilder.create().build();
                    HttpResponse res = client.execute(post);
                    String s = res.getStatusLine().toString();
                    if(s.equals("HTTP/1.1 200 OK")) {
                        System.out.println("Navigated");
                        //your code to handle logic 
                    } else {
                        //your code to handle logic with other response code
                    }
                }
            }
        } catch (Exception e) {
            System.out.println(e.getStackTrace());
        }
    }    
}

score 0 · Answer 7 · answered Apr 17 '20 at 15:21

Compacting the last answers into a few lines:

In Python:

# extract link elements then extract the href itself.
# browser is the driver opened with the web page

link_elements = browser.find_elements_by_xpath('//*[@class="child-category-container"]/a')
for link in link_elements:        
    url = link.get_attribute('href')
    urls.append(url)

score 0 · Answer 8 · answered Sep 23 '20 at 14:51

Not sure how efficient it is but I reloaded the links in the same list after each iteration and successfully completed the task.

String baseURL = "https://www.wikipedia.org/";

        driver.get(baseURL);
        List<WebElement> links = driver.findElements(By.xpath("//div[@class='central-featured']/div/a")); 
        System.out.println("The size of the list is: " + links.size());

        // Loop through links, click on each link, navigate back, reload the link and
        // continue.

        for (int i = 0; i < links.size(); ++i) {
            links.get(i).click();
            driver.navigate().back();
            // reloading the list or there will be stale-element exception
            links = driver.findElements(By.xpath("//div[@class='central-featured']/div/a"));

        }
        // print the link text and href values

        for (int i = 0; i < links.size(); ++i) {
            System.out.print(links.get(i).getText() + "--> " + links.get(i).getAttribute("href"));
        }

        driver.close();

score 0 · Answer 9 · edited Mar 06 '22 at 17:02

0

you can use something like this

list_links = driver.find_elements_by_tag_name('a')

for i in list_links:
        print(i.get_attribute('href'))

driver.quit()

edited Mar 06 '22 at 17:02

Wahyu Kristianto

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6
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answered Mar 06 '22 at 16:54

Charan Sai

1

How to click on All links in web page in Selenium WebDriver

9 Answers9

Java

Python

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