Is it possible to UI test Window's Menu ? ```onNod...

# compose-desktop

[Olivier Patry]

Is it possible to UI test Window's Menu ?

onNodeWithText("Tools").assertIsDisplayed()

This fails ("Tools" being one of the MenuBar's root Menu).
I'm running on macOS if that matters (given how the macOS menu are managed vs app Window, it could)

(See code in thread 🧵)

[Olivier Patry]

@Test
fun `when clicking on show network logs then should trigger callback`() = runComposeUiTest {
    var isNetworkLogClicked = false
    setContent {
        Window(
            onCloseRequest = {},
            title = "Test",
        ) {
            AppMenuBar(
                onNetworkLogClick = { isNetworkLogClicked = true }
            )
        }
    }

    onNodeWithText("Tools")
        .assertIsDisplayed()
        .performClick()

    onNodeWithText("Show network logs")
        .assertIsDisplayed()
        .performClick()
}

[Olivier Patry]

@Composable
fun FrameWindowScope.AppMenuBar(
    onNetworkLogClick: () -> Unit,
) {
    MenuBar {
        Menu("Tools", mnemonic = 'T') {
            Item(
                text = "Show network logs",
                shortcut = when (currentOS) {
                    OS.Mac -> KeyShortcut(key = Key.L, meta = true, shift = true)
                    OS.Linux,
                    OS.Windows -> KeyShortcut(key = Key.L, ctrl = true, shift = true)
                },
                onClick = onNetworkLogClick
            )
        }
    }
}

[Alexander Maryanovsky]

Currently no. UI test doesn’t create a real window, and even if it did, on macOS the toolbar isn’t part of the node tree.

[Olivier Patry]

That's what I thought.
I hoped for a way to invoke the composable in dedicated context just to stress it's behavior (callbacks)