eliminating duplicate links on the webpage and avoid link is stale error

Question

I have a list of 20 links and some of them are duplicates. I click onto the first link which leads me to the next page, I download some files from the next page.

Page 1

• Link 1
• Link 2
• Link 3
• link 1
• link 3
• link 4
• link 2

Link 1 (click) --> (opens) Page 2

Page 2 (click back button browser) --> (goes back to) Page 1

Now I click on Link 2 and repeat the same thing.

             System.setProperty("webdriver.chrome.driver", "C:\\chromedriver.exe"); 
    String fileDownloadPath = "C:\\Users\\Public\\Downloads"; 


    //Set properties to supress popups
    Map<String, Object> prefsMap = new HashMap<String, Object>();
    prefsMap.put("profile.default_content_settings.popups", 0);
    prefsMap.put("download.default_directory", fileDownloadPath);
    prefsMap.put("plugins.always_open_pdf_externally", true);
    prefsMap.put("safebrowsing.enabled", "false"); 

    //assign driver properties
    ChromeOptions option = new ChromeOptions();
    option.setExperimentalOption("prefs", prefsMap);
    option.addArguments("--test-type");
    option.addArguments("--disable-extensions");
    option.addArguments("--safebrowsing-disable-download-protection");
    option.addArguments("--safebrowsing-disable-extension-blacklist");


    WebDriver driver  = new ChromeDriver(option);
           driver.get("http://www.mywebpage.com/");

           List<WebElement> listOfLinks = driver.findElements(By.xpath("//a[contains(@href,'Link')]"));
        Thread.sleep(500);



        pageSize = listOfLinks.size();

        System.out.println( "The number of links in the page is: " + pageSize);

        //iterate through all the links on the page
        for ( int i = 0; i < pageSize; i++)
        {

            System.out.println( "Clicking on link: " + i );
            try 
            {
                    linkText = listOfLinks.get(i).getText();
                    listOfLinks.get(i).click();
            }
            catch(org.openqa.selenium.StaleElementReferenceException ex)
            {
                listOfLinks = driver.findElements(By.xpath("//a[contains(@href,'Link')]"));
                linkText = listOfLinks.get(i).getText();
                listOfLinks.get(i).click();
            }
               try 
            {
              driver.findElement(By.xpath("//span[contains(@title,'download')]")).click();

            }
            catch (org.openqa.selenium.NoSuchElementException ee)
            {
                driver.navigate().back();
                Thread.sleep(300);
                continue;
            }
      Thread.sleep(300);                 
            driver.navigate().back();
            Thread.sleep(100);
        }

The code is working fine and clicks on all the links and downloads the files. Now I need to improve the logic omit the duplicate links. I tried to filter out the duplicates in the list but then not sure how should I handle the org.openqa.selenium.StaleElementReferenceException. The solution I am looking for is to click on the first occurrence of the link and avoid clicking on the link if it re-occurs.

(This is part of a complex logic to download multiple files from a portal >that I don't have control over. Hence please don't come back with the >questions like why there are duplicate links on the page at the first place.)

Answer 1

First I don't suggest you to be doing requests (findElements) to the WebDriver repeatedly, you will see a lot of performance issues following this path, mainly if you have a lot of links, and pages.

Also if you are doing the same thing always on the same tab, you will need to wait the refresh 2 times ( page of the links and page of the download ), now if you open each link in a new tab, you just need to wait the refresh of the page where you will download.

I have a suggestion, just distinct repeated links as @supputuri said and open each link in a NEW tab, in this way you don't need to handle stale, don't need to be searching on the screen every time for the links and don't need to wait the refresh of the page with links in each iteration.

List<WebElement> uniqueLinks = driver.findElements(By.xpath("//a[contains(@href,'Link')][not(@href = following::a/@href)]"));

for ( int i = 0; i < uniqueLinks.size(); i++)
{
    new Actions(driver)
         .keyDown(Keys.CONTROL)
         .click(uniqueLinks.get(i))
         .keyUp(Keys.CONTROL)
         .build()
         .perform();
    // if you want you can create the array here on this line instead of create inside the method below.
    driver.switchTo().window(new ArrayList<>(driver.getWindowHandles()).get(1));
    //do your wait stuff.
    driver.findElement(By.xpath("//span[contains(@title,'download')]")).click();
    //do your wait stuff.
    driver.close();
    driver.switchTo().window(new ArrayList<>(driver.getWindowHandles()).get(0));
}

I'm not in a place where I was able to test my code properly right now, any issues on this code just comment and I will update the answer, but the idea is right and it's pretty simple.

Answer 2

First lets see the xpath.

Sample HTML:

<!DOCTYPE html>
<html>
 <body>
 <div>
  <a href='https://google.com'>Google</a>
  <a href='https://yahoo.com'>Yahoo</a>
  <a href='https://google.com'>Google</a>
  <a href='https://msn.com'>MSN</a>
 </body>
</html>

Let's see the xpath to get the distinct Links out of the above.

//a[not(@href = following::a/@href)]

The logic in xpath is we are making sure the href of the link is not matching with any following links href, if it's match then it's considered as duplicate and xpath does not return that element.

Stale Element: So, now it's time to handle the stale element issue in your code. The moment you click on the Link 1 all the references stored in listOfLinks will be invalid as selenium will get assign the new references to the elements each time they load on the page. And when you try to access the elements with old reference you will get the stale element exception. Here is the snippet of code that should give you an idea.

List<WebElement> listOfLinks = driver.findElements(By.xpath("//a[contains(@href,'Link')][not(@href = following::a/@href)]"));
Thread.sleep(500);
pageSize = listOfLinks.size();
System.out.println( "The number of links in the page is: " + pageSize);
//iterate through all the links on the page
for ( int i = 0; i < pageSize; i++)
{
    // ===> consider adding step to explicit wait for the Link element with "//a[contains(@href,'Link')][not(@href = following::a/@href)]" xpath present using WebDriverWait 
    // don't hard code the sleep 
    // ===> added this line
    <WebElement> link = driver.findElements(By.xpath("//a[contains(@href,'Link')][not(@href = following::a/@href)]")).get(i);
    System.out.println( "Clicking on link: " + i );
    // ===> updated next 2 lines
    linkText = link.getText();
    link.click();
    // ===> consider adding explicit wait using WebDriverWait to make sure the span exist before clicking. 
    driver.findElement(By.xpath("//span[contains(@title,'download')]")).click();
    // ===> check this answer (https://stackoverflow.com/questions/34548041/selenium-give-file-name-when-downloading/56570364#56570364) for make sure the download is completed before clicking on browser back rather than sleep for x seconds.
    driver.navigate().back();
    // ===>  removed hard coded wait time (sleep)
}

xpath ScreenShot:

Edit1:

If you want to open the link in the new window then use the below logic.

WebDriverWait wait = new WebDriverWait(driver, 20);
        wait.until(ExpectedConditions.presenceOfAllElementsLocatedBy(By.xpath("//a[contains(@href,'Link')][not(@href = following::a/@href)]")));
        List<WebElement> listOfLinks = driver.findElements(By.xpath("//a[contains(@href,'Link')][not(@href = following::a/@href)]"));
        JavascriptExecutor js = (JavascriptExecutor) driver; 
        for (WebElement link : listOfLinks) {
            // get the href
            String href = link.getAttribute("href");
            // open the link in new tab
            js.executeScript("window.open('" + href +"')");
            // switch to new tab
            ArrayList<String> tabs = new ArrayList<String> (driver.getWindowHandles());
            driver.switchTo().window(tabs.get(1));
            //click on download

            //close the new tab
            driver.close();
            // switch to parent window
            driver.switchTo().window(tabs.get(0));
         }

Screenshot: Sorry for the poor quality of the screenshot, could not upload the high quality video due to size limit.

Answer 3

you can do like this.

1. Save Index of element in the list to a hashtable
2. if Hashtable already contains, skip it
3. once done, HT has only unique elements, ie first foundones

4. Values of HT are the index from listOfLinks

           HashTable < String, Integer > hs1 = new HashTable(String, Integer);
           for (int i = 0; i < listOfLinks.size(); i++) {
               if (!hs1.contains(e.getText()) {
   
                       hs1.add(e.getText(), i);
                   }
               }
               for (int i: hs1.values()) {
   
                   listOfLinks.get(i).click();
               }