问题描述
点击视图A有一个按钮，我们去视图B
View A has a button upon clicking it, we go to view B
- 视图 B 不保留指向视图 A 的指针.
- 从视图 B，我想(以编程方式)加载回视图 A
实际上，我想杀死 B 并用 A 替换它.
Effectively, i'd like to kill B and replace it with A.
我认为以下应该有效，但没有
I was thinking that the following should work but, it does not
从视图 B 调用
ViewController *main = [ViewController new];
[self addSubview:[main view]];
请问我遗漏了什么?
推荐答案
我个人认为最简单的方法是让 UIViewController 和一个 IBOutlet两个 UIView 对象.您可以在界面构建器中添加和设计它们，只需将其中一个(视图 B)设置为隐藏(它是 UIView 中的一个属性).然后，您可以指定一个按钮操作来切换视图 B 的可见性.
Personally I think the easiest way to do this would be by having a UIViewController with an IBOutlet to both UIView objects. You can add and design them both in the interface builder and just set one of them (view B) as hidden (it's a property in UIView).Then, you could specify a button action to toggle the visibility of view B.
我必须补充一点，有一些用于实现屏幕流的结构，例如 NavigationController.但是，在您的情况下，您也可以考虑使用 presentModalViewController:animated: 方法.
I must add though that there are constructs for implementing screen flows, such as the NavigationController. In your case, however, you might also consider the use of the presentModalViewController:animated: method.
这完全取决于实际情况，但一般来说，更好的做法是为应用程序中的每个 UIView 制作一个单独的 UIViewController.
It all depends really, but in general it's better practice to make a seperate UIViewController for each UIView in your application.
希望这会有所帮助！
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