Question

如何触摸UITests的特定坐标？

当我在特定地点录制时，我有类似的内容：

XCUIApplication *app = [[XCUIApplication alloc] init];
[[[[[[[[[[app.otherElements containingType:XCUIElementTypeNavigationBar 
identifier:@"Navigation Bar title"] 
childrenMatchingType:XCUIElementTypeOther].element 
childrenMatchingType:XCUIElementTypeOther].element 
childrenMatchingType:XCUIElementTypeOther].element 
childrenMatchingType:XCUIElementTypeOther] elementBoundByIndex:0] 
childrenMatchingType:XCUIElementTypeOther].element 
childrenMatchingType:XCUIElementTypeOther] 
elementBoundByIndex:0].staticTexts[@"Action"] tap];

Answer 1

您只能点击已知元素引用的特定坐标。意思是，您无法点击坐标（20,400）处的像素。相反，您需要找到一个元素，然后点击带有偏移量的东西。

XCUIApplication *app = [[XCUIApplication alloc] init];
XCUIElement *label = app.labels[@"Label Name"];
XCUICoordinate *coordinate = [label coordinateWithNormalizedOffset(CGVectorMake(0.5, 1.2));
[coordinate tap];

我记录了有关如何在UI Testing Cheat Sheet帖子中找出正确偏移量的更多信息。

如果您只是想点击Action按钮，则可以直接访问它（而不是深入查看所有这些查询）。

XCUIApplication *app = [[XCUIApplication alloc] init];
[[[app.navigationBars element].staticTexts[@"Action"] tap];

Answer 2

要点按屏幕的特定坐标，请从窗口元素构建XCUICoordinate。

XCUIApplication *app = [[XCUIApplication alloc] init];
XCUIElement *window = [app.windows elementAtIndex:0];
// Get co-ordinate for the top left corner of the screen
XCUICoordinate *origin = [window coordinateWithNormalizedOffset:CGVectorMake(0.0, 0.0)];
// Get coordinate relative to the top left of the screen
XCUICoordinate *myCoordinate = [origin coordinateWithOffset:CGVectorMake(40.0, 100.0)];

[coordinate tap];

Answer 3

它是快速版本

let point = app.labels["labelName"].coordinate(withNormalizedOffset: CGVector(dx: 0, dy:0))
point.tap()

触摸特定坐标（UI测试）

3 个答案: