在Swift中以编程方式进行Segue [英] Make Segue programmatically in Swift

问题描述

我有两个VC：VC1和VC2。
在VC1中，我有一个完成按钮我以编程方式制作了一个结果数组我想通过到VC2。

I have two VCs: VC1 and VC2. In VC1, I have a finish button which I programmatically made and a result array I want to pass to VC2.

我知道如何在故事板中制作Segue，但是由于完成按钮以编程方式制作。

I know how to make Segue in Storyboard, but I cannot do that at this moment since the finish button is programmatically made.

如果我想使用segue传递结果数组，有没有办法以编程方式生成segue？如果这不可能，我应该使用 presentViewController 来呈现VC2并设置委托来传递结果数组？

If I want to pass result array using segue, is there a way to make the segue programmatically? If this is not possible, should I just present VC2 using presentViewController and set a delegate to pass the result array?

推荐答案

你可以像在这个答案中提出的那样做：
InstantiateViewControllerWithIdentifier 。

You can do it like proposed in this answer: InstantiateViewControllerWithIdentifier.

此外我正在为您提供代码来自Swift中重写的链接答案，因为链接中的答案最初是用Objective-C编写的。

Furthermore I am providing you the code from the linked answer rewritten in Swift because the answer in the link was originally written in Objective-C.

let vc = UIStoryboard(name:"Main", bundle:nil).instantiateViewControllerWithIdentifier("identifier") as! SecondViewController

vc.resultsArray = self.resultsArray
self.navigationController?.pushViewController(vc, animated:true)

编辑：

由于这个答案引起了一些关注，我想我还会为您提供更多故障安全方式。在上面的回答中，如果带有identifier的 ViewController 不是类型 SecondViewController 。在Swift中，您可以通过使用可选绑定来防止此崩溃：

Since this answer draws some attention I thought I provide you with another more failsafe way. In the above answer the application will crash if the ViewController with "identifier" is not of type SecondViewController. In Swift you can prevent this crash by using optional binding:

guard let vc = UIStoryboard(name:"Main", bundle:nil).instantiateViewControllerWithIdentifier("identifier") as? SecondViewController else {
    print("Could not instantiate view controller with identifier of type SecondViewController")
    return
}

vc.resultsArray = self.resultsArray
self.navigationController?.pushViewController(vc, animated:true)

这样 ViewController 如果类型为 SecondViewController ，则被推送。如果无法转换为 SecondViewController ，则会打印一条消息，并且应用程序仍保留在当前 ViewController 上。

This way the ViewController is pushed if it is of type SecondViewController. If can not be casted to SecondViewController a message is printed and the application remains on the current ViewController.

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