How to perform search on iOS device with XCUITest

Support
1

Hello,
I am working on automation testing on real iOS devices with Appium driver and XCUITest automator.
What I want my test to do is to enter text in a search bar and then to press search button.
However, I can not find a solution how to execute the searching action.
What I use for Android is await driver.execute(‘mobile: performEditorAction’, {action: ‘search’});, but I can not find similar action for iOS.
Could you please advise what would be the appropriate action for iOS?
Thank you in advance!

2

https://appiumpro.com/editions/8-how-to-find-elements-in-ios-not-by-xpath

3

Thanks for the link. Definitely will use it for other questions, but for the current one I do not see any information.

4

AFAIK there is no direct alternative in iOS. I would try to locate the corresponding button on the on-screen keyboard and tap it

5

You mean last button on keyboard which changes on iOS depending on behavior: Next, Done, Go, Search and so on?

6

Yeah, this one. Maybe I could try with commands, which are for “Enter” button, like this one: driver.getKeyboard().pressKey(Keys.ENTER) . However, I tried this …getKeyboard() function, but it was not recognized in my project environment. It is strange how XCUITest does not support such common activities Search, Enter…etc.

7

Yeah, I will try this one, nevertheless it seems foolish approach, because as I said before, searching is a often used function, which obviously is not supported. :expressionless:

8

In the end, I will use the approach await $(’…’).setvalue(“some text” + “\n”). This will enter the “some text” in the search field with locator(selector) $(’…’) and the ‘\n’ will be executed as “Enter”. Another option to separate entering the text in the field and ‘pressing’ Enter is using these 2 function one after another:

  1. await $(’…’).setValue(“some text”);
  2. await $(’…’).addValue(’\n’);
9

finds last key on open keyboard for iOS

iOSClassChain = "**/XCUIElementTypeKeyboard[1]/**/XCUIElementTypeButton[-1]"
1 Like
10

Definitely will try this. Thanks!