Shammi Shammi - 1 year ago 92
Java Question

How to fetch all links and click those links one by one using selenium webdriver

I using selenuim webdriver, lang: java
I am fetching all links from webpage and trying to click each link one by one. I am getting below error:

error org.openqa.selenium.StaleElementReferenceException: Element not found in the cache - perhaps the page has changed since it was looked up
Command duration or timeout: 30.01 seconds
For documentation on this error, please visit:
Build info: version: '2.25.0', revision: '17482', time: '2012-07-18 21:09:54'

and here is my code :

public void getLinks()throws Exception{
try {
List<WebElement> links = driver.findElements(By.tagName("a"));
int linkcount = links.size();
for (WebElement myElement : links){
String link = myElement.getText();
if (link !=""){;
}catch (Exception e){
System.out.println("error "+e);

actually, it's displaying in output

[[FirefoxDriver: firefox on XP (ce0da229-f77b-4fb8-b017-df517845fa78)] -> tag name: a]

as link, I want to eliminate these form result.

Answer Source

There is no such a good idea to have following scenario :

for (WebElement element : webDriver.findElements(locator.getBy())){;

Why? Because there is no guarantee that the; will have no effect on other found elements, so the DOM may be changed, so hence the StaleElementReferenceException.

It is better to use the following scenario :

int numberOfElementsFound = getNumberOfElementsFound(locator);
for (int pos = 0; pos < numberOfElementsFound; pos++) {
  getElementWithIndex(locator, pos).click();

This is better because you will always take the WebElement refreshed, even the previous click had some effects on it.

EDIT : Example added

  public int getNumberOfElementsFound(By by) {
    return webDriver.findElements(by).size();

  public WebElement getElementWithIndex(By by, int pos) {
    return webDriver.findElements(by).get(pos);

Hope to be enough.

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