How can I find element by non-unique resousre-id?

Question

I test an application which use non-unique resourse-id for elements.

Is there any way to find such elements by xpath like

//*[@resourse-id='non-unique-id'][2]

I mean the second element with same resourse-id.

Answer 1

I'd recommend avoiding xpath in mobile automation since this is the most time-consuming strategy to find elements. If you don't have any other anchors for your elements but you confident in its order, you can stick to the following approach: Appium driver can return a list of elements with the same locator, in case of Page Object model you can either do this way:

@AndroidFindBy(uiAutomator = "resourceIdMatches(\".*whatever\")")
private List<MobileElement> elements;

so, once your page is initialized, you can access an element by index:

elements.get(1).click();

or, in case of manual managenemt, you can do this way:

List<MobileElement> elements = driver.findElements(MobileBy.AndroidUIAutomator("resoureceIdMatches(\".*whatever\")"));
elements.get(3).click();

Hope this helps.

Answer 2

As far as my understanding goes, you need to select the second element with the path as mentioned: //*[@resourse-id='non-unique-id']

To do that, you need to first grab all the elements with the same non-unique resource ID and then get() them. So, your code should be:

driver.findElements(By.xpath("//*[@resourse-id='non-unique-id']")).get(1).click();

The index for any list starts at 0. So, the second element can be accessed through the value of 1.

Hope this helps.

Answer 3

Try following approach:

(//*[@resourse-id='non-unique-id'])[2]

Answer 4

HTML with non-unique ids is not a valid HTML document.

So, for the sake of future testability, ask the developers to fix the ids.