为什么我无法使用 UI 自动化脚本访问 iPhone 应用程序中的后退按钮？

Question

我有一个简单的函数应该检查视图是否在主界面，如果不是，将其带到主界面：

function returnHome() {
    if (UIATarget.localTarget().frontMostApp().navigationBar().name() == mainTitle) return true;

    // set tab bar to calculations
    UIALogger.logMessage("Set tab bar to main.");
    if (UIATarget.localTarget().frontMostApp().tabBar().selectedButton().name() != mainTabName) {
        UIATarget.localTarget().frontMostApp().tabBar().buttons()[mainTabName].tap();
    }

    // go back to the home
    UIALogger.logMessage("Go back to home.");
    var backButton = UIATarget.localTarget().frontMostApp().mainWindow().buttons()["Back"];
    if (backButton.isValid()) {
        backButton.tap();
    } else {
        UIALogger.logError("Could not find 'Back' button!");
    }
}

现在，我无法通过错误“找不到‘返回’按钮！”。它就在上面，在左角，正盯着我，嘲弄我！

UIATarget.localTarget().frontMostApp().logElementTree() 给了我以下树：

2) UIAApplication [name:MyApplication value:(null) rect:{{x:0, y:20}, {width:320, height:460}}]
3) UIAWindow [name:(null) value:(null) rect:{{x:0, y:0}, {width:320, height:480}}]
4) UIAImage [name:(null) value:(null) rect:{{x:0, y:108}, {width:320, height:367}}]
4) UIANavigationBar [name:(null) value:(null) rect:{{x:0, y:64}, {width:320, height:44}}]
5) UIASegmentedControl [name:(null) value:(null) rect:{{x:5, y:71}, {width:310, height:30}}]
6) UIAButton [name:SementedControllerTagC, 3 of 3 value:(null) rect:{{x:212, y:71}, {width:103, height:30}}]
6) UIAButton [name:SementedControllerTagB, 2 of 3 value:(null) rect:{{x:109, y:71}, {width:102, height:30}}]
6) UIAButton [name:SementedControllerTagA, 1 of 3 value:1 rect:{{x:5, y:71}, {width:103, height:30}}]
4) UIATableView [name:Empty list value:rows 1 to 2 of 3 rect:{{x:15, y:169}, {width:290, height:247}}]
5) UIAButton [name:XYZ value:(null) rect:{{x:20, y:121}, {width:280, height:46}}]
5) UIATableGroup [name:SomeText value:(null) rect:{{x:15, y:169}, {width:290, height:50}}]
6) UIAStaticText [name:SomeText value:(null) rect:{{x:25, y:197}, {width:270, height:22}}]
5) UIATableCell [name:A value:(null) rect:{{x:15, y:219}, {width:290, height:50}}]
6) UIAElement [name:B value:(null) rect:{{x:15, y:219}, {width:247, height:50}}]
6) UIAButton [name:more info, A value:(null) rect:{{x:262, y:219}, {width:43, height:50}}]
5) UIATableCell [name:X value:(null) rect:{{x:15, y:269}, {width:290, height:50}}]
6) UIAElement [name:Y value:(null) rect:{{x:15, y:269}, {width:247, height:50}}]
6) UIAButton [name:more info, B value:(null) rect:{{x:262, y:269}, {width:43, height:50}}]
4) UIAButton [name:(null) value:(null) rect:{{x:85, y:339}, {width:150, height:32}}]
4) UIAImage [name:(null) value:(null) rect:{{x:0, y:334}, {width:320, height:97}}]
4) UIAStaticText [name:SomeText value:(null) rect:{{x:45, y:377}, {width:189, height:21}}]
4) UIAStaticText [name:SomeText value:(null) rect:{{x:45, y:406}, {width:230, height:21}}]
4) UIAStaticText [name:SomeText value:(null) rect:{{x:234, y:377}, {width:41, height:21}}]
4) UIAStaticText [name:(null) value:(null) rect:{{x:0, y:0}, {width:0, height:21}}]
4) UIAStaticText [name:(null) value:(null) rect:{{x:45, y:377}, {width:230, height:21}}]
4) UIAStaticText [name:(null) value:(null) rect:{{x:45, y:406}, {width:230, height:21}}]
4) UIANavigationBar [name:mainNavigationBar value:(null) rect:{{x:0, y:20}, {width:320, height:44}}]
5) UIAStaticText [name:mainNavigationBar value:(null) rect:{{x:108, y:28}, {width:105, height:27}}]
5) UIAButton [name:Back value:(null) rect:{{x:5, y:27}, {width:49, height:30}}]
4) UIATabBar [name:(null) value:(null) rect:{{x:0, y:431}, {width:320, height:49}}]
5) UIAButton [name:tab1 value:1 rect:{{x:2, y:432}, {width:103, height:48}}]
5) UIAButton [name:tab2 value:(null) rect:{{x:109, y:432}, {width:102, height:48}}]
5) UIAButton [name:tab3 value:(null) rect:{{x:215, y:432}, {width:103, height:48}}]

日志清楚地显示了“返回”按钮，但我没有得到正确的参考。谁能在这里指出我正确的方向？难道我做错了什么？仅供参考，我尝试从frontMostApp()、mainWindow() 和navigationBar() 访问buttons()。都失败了。

Answer 1

“后退”按钮不是窗口的直接子窗口，而是导航栏。尝试使用以下代码访问它：

   var backButton = UIATarget.localTarget().frontMostApp().mainWindow().navigationBar().buttons()["Back"]; 

Answer 2

您是如何构建视图层次结构的？从日志来看，您在一个 UIWindow 下似乎有两个导航栏：一个未标记，一个标记有 mainNavigationBar。

正如克劳斯指出的那样，您应该能够通过访问主窗口的导航栏来访问后退按钮，但在您的情况下，由于存在第二个导航栏，这可能会失败。

如果有帮助，以下是我编写的脚本，它逐步遍历 UICatalog 示例应用程序中的每个部分，抓取该部分的屏幕截图，然后返回导航控制器的主级别。在此示例中，我使用leftButton() 访问导航栏上的后退按钮。

var target = UIATarget.localTarget();
var application = target.frontMostApp(); 
var mainWindow = application.mainWindow();
var tableView = mainWindow.tableViews()[0];


UIALogger.logMessage("Number of cells " + tableView.cells().length);

for (var currentCellIndex = 0; currentCellIndex < tableView.cells().length; currentCellIndex++)
{
    var currentCell = tableView.cells()[currentCellIndex];
    UIALogger.logStart("Testing table option " + currentCell.name());

    currentCell.scrollToVisible();  

    target.delay(1);
    currentCell.tap();  // Go down a level
    target.delay(1);

    UIATarget.localTarget().captureScreenWithName(currentCell.name());
    mainWindow.navigationBar().leftButton().tap(); // Go back
    target.delay(1);
    UIALogger.logPass("Testing table option " + currentCell.name());
}

Answer 3

也许它不是一个真正的按钮。我遇到了类似的问题，日志将它作为按钮返回，但它实际上是一个视图。因此，我的解决方案是找到项目（可以是图像），然后获取 rect(x, y)，然后在屏幕上使用特定位置的选项卡，例如 target.tapWithOptions({x:5, y:190});希望这会有所帮助，干杯