如何在Safari中打开url，并在Xcode 7中的UITests下返回到应用程序? [英] How to open url in Safari and the get back to the app under UITests in Xcode 7?

问题描述

这是我的自定义视图，其中"LondonStreet"是一个按钮.

This is my custom view where "LondonStreet" is a button.

当我点击该按钮时，我会得到url并在Safari中打开它(它有效).然后，我可以使用"Back to Wishlist"按钮返回它(也可以使用).

When I tap that button I get url and open it in Safari (it works). Then I can go back, using "Back to Wishlist" button (it also works).

问题是当我尝试在UITests下进行测试时.

The problem is when I try to test this under UITests.

itemsTable.cells.elementBoundByIndex(0).buttons["addressButton"].tap() //press the button to open link in Safari

沿这行:

app.statusBars.buttons["Back to Wishlist"].tap() //go back, doesn't work, although it was recorded by Xcode itself.

是错误:

UI测试失败-无法在5秒内获取屏幕截图.

UI Testing Failure - Failed to get screenshot within 5s.

以及 issue Navigator

UI测试失败-无法及时更新应用程序状态.

UI Testing failure - Unable to update application state promptly.

推荐答案

从iOS 11开始，您可以使用XCUIApplication(bundleIdentifier :)初始化程序与其他应用程序进行交互.

Starting in iOS 11 you can interact with other applications using the XCUIApplication(bundleIdentifier:) initializer.

要返回到您的应用，您可以执行以下操作:

To get back to your app you'd do something like:

let myApp = XCUIApplication(bundleIdentifier: "my.app.bundle.id")
let safari = XCUIApplication(bundleIdentifier: "com.apple.mobilesafari")

// Perform action in your app that opens Safari

safari.wait(for: .runningForeground, timeout: 30)
myApp.activate() // <--- Go back to your app

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