Question

在Obj-c中我做了一个switch语句，我曾经在我的应用程序中使用UIsplitviewcontroller在iPad中移动现在我想在swift中做同样的事情...我尝试了几个小时，现在我唯一能够尝试的是代码因为它说某种编译错误无论如何这是我在Obj-c中得到的东西

-(void)initialSite:(int)viewId {
UIViewController *viewController;
switch (viewId) {
    case 0:{
        viewController = self.initital;
        NSString *star = [NSString stringWithFormat:@"Velkommen til %@'s  Bog",[data valueForKey:@"navn"]];
        self.navigationItem.title = star;}
        break;
    case 1:{
        viewController = self.startSide;
        NSString *start = [NSString stringWithFormat:@"%@'s  Bog, start-side",[data valueForKey:@"navn"]];
        self.navigationItem.title = start;}
        break;
}
[self showChildViewController:viewController];

}

这是我迄今为止在swift中提出的问题。仍然是新手，并了解它有点难，即使我有快速的编程语言书

这是我到目前为止所获得的快速

    let viewController = UIViewController()
switch viewController {
case "initial":
let initial : UIStoryboard = UIStoryboard(name: "Main", bundle: nil)
let vc0 : UIViewController = initial.instantiateViewControllerWithIdentifier("initial") as UIViewController
self.presentViewController(vc0, animated: true, completion: nil)
let rowData: NSDictionary = self.menuItemArray[indexPath.row] as NSDictionary!
self.navigation.title = rowData["navn"] as? String

case "startSide":
let startSide : UIStoryboard = UIStoryboard(name: "Main", bundle: nil)
let vc1 : UIViewController = startSide.instantiateViewControllerWithIdentifier("startSide") as UIViewController
let rowData: NSDictionary = self.manuItemArray[indexPath.row] as NSDictionary!
self.presentViewController(vc1, animated: true, completion: nil)
self.navigation.title = rowData["navn"] as? String
default:


}

错误是： 预期声明 让viewController = UIViewcontroller（）

Answer 1

让我们从您的Obj-C实施开始：

-(void)initialSite:(int)viewId 
{
    UIViewController *viewController;
    switch (viewId) 
    {
        case 0:
        {
            viewController = self.initital;
            NSString *star = [NSString stringWithFormat:@"Velkommen til %@'s  Bog",[data valueForKey:@"navn"]];
            self.navigationItem.title = star;
        }
        break;
        case 1:
        {
            viewController = self.startSide;
            NSString *start = [NSString stringWithFormat:@"%@'s  Bog, start-side",[data valueForKey:@"navn"]];
            self.navigationItem.title = start;
        }
        break;
    }

    [self showChildViewController:viewController];
}

现在Swift中的这个片段：

func initialSite(viewID:Int)
{
    var viewController : UIViewController?

    switch (viewID)
    {
        case 0:
            viewController = self.initial
            let navn = self.data["navn"] as? String
            let star = "Velkommen til \(navn)'s Bog"
            self.navigationItem.title = star

        case 1:
            viewController = self.startSide
            let navn = self.data["navn"] as? String
            let star = "\(navn)'s Bog, start-side"
            self.navigationItem.title = star

        default:
            viewController = nil
            // do nothing
    }

    self.showChildViewController(viewController)
}

您必须记住的主要事项是var与let的区别。通常情况下，您会使用let来创建内容，除非这些内容的值稍后会更改，您使用var。

另一件事是使用带有?后缀的选项。这是当值可能是nil（未设置）时，否则它必须包含值。

Answer 2

看起来SiLo打败了我。无论如何，我有我的解决方案，所以我会发布它。我就是这样做的：

func initialSite(viewId: Int) -> () {
    var viewController: UIViewController?
    let dataValue = data["navn"];
    var start: String?

    switch viewId {
        case 1:
            viewController = self.initital
            start = "Velkommen til \(dataValue)'s  Bog"
        case 2:
            viewController = self.startSide
            start = "\(dataValue)'s  Bog, start-side"
        default:
            break;
    }

    self.navigationItem.title = start!
    showChildViewController(viewController!)
}

切换语句出错 - 语言迅速

2 个答案: