我在UIViewController上实现了一个类库，简化了这个操作。 基本上，你在NSDictionary中设置你想要传递的参数与执行segue的uiitem关联。它也适用于手动segue。

例如，你可以这样做

[self performSegueWithIdentifier:@"yourIdentifier" parameters:@{@"customParam1":customValue1, @"customValue2":customValue2}];

对于手动segue或者创建一个带有segue的按钮并使用

[button setSegueParameters:@{@"customParam1":customValue1, @"customValue2":customValue2}];

如果目标视图控制器不符合键的键值编码，什么都不会发生。它也适用于键值(对unwind segue很有用)。 点击这里查看 https://github.com/stefanomondino/SMQuickSegue

我的解决方案是类似的。

// In destination class: 
var AddressString:String = String()

// In segue:
override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
   if (segue.identifier == "seguetobiddetailpagefromleadbidder")
    {
        let secondViewController = segue.destinationViewController as! BidDetailPage
        secondViewController.AddressString = pr.address as String
    }
}

我使用这个解决方案，这样我就可以在同一个函数中保持对segue的调用和数据通信:

private var segueCompletion : ((UIStoryboardSegue, Any?) -> Void)?

func performSegue(withIdentifier identifier: String, sender: Any?, completion: @escaping (UIStoryboardSegue, Any?) -> Void) {
    self.segueCompletion = completion;
    self.performSegue(withIdentifier: identifier, sender: sender);
    self.segueCompletion = nil
}

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    self.segueCompletion?(segue, sender)
}

用例应该是这样的:

func showData(id : Int){
    someService.loadSomeData(id: id) {
        data in
        self.performSegue(withIdentifier: "showData", sender: self) {
            storyboard, sender in
            let dataView = storyboard.destination as! DataView
            dataView.data = data
        }
    }
}

这似乎对我有用，但是，我不能100%确定perform和prepare函数总是在同一个线程上执行。

只需要在prepareForSegue:方法中获取一个对目标视图控制器的引用，并将你需要的任何对象传递到那里。这里有一个例子……

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    // Make sure your segue name in storyboard is the same as this line
    if ([[segue identifier] isEqualToString:@"YOUR_SEGUE_NAME_HERE"])
    {
        // Get reference to the destination view controller
        YourViewController *vc = [segue destinationViewController];

        // Pass any objects to the view controller here, like...
        [vc setMyObjectHere:object];
    }
}

修改:您还可以使用performSegueWithIdentifier:sender:方法来激活基于选择或按钮按下的转换到新视图。

例如，假设我有两个视图控制器。第一个包含三个按钮，第二个需要知道在转换之前按下了哪个按钮。你可以在代码中使用performSegueWithIdentifier:方法将按钮连接到IBAction，就像这样…

// When any of my buttons are pressed, push the next view
- (IBAction)buttonPressed:(id)sender
{
    [self performSegueWithIdentifier:@"MySegue" sender:sender];
}

// This will get called too before the view appears
- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    if ([[segue identifier] isEqualToString:@"MySegue"]) {

        // Get destination view
        SecondView *vc = [segue destinationViewController];

        // Get button tag number (or do whatever you need to do here, based on your object
        NSInteger tagIndex = [(UIButton *)sender tag];

        // Pass the information to your destination view
        [vc setSelectedButton:tagIndex];
    }
}

编辑:我最初附上的演示应用程序现在已经有6年了，所以我删除了它以避免任何混淆。

有时避免在两个视图控制器之间创建编译时依赖关系是有帮助的。下面是你不用关心目标视图控制器的类型就能做到的方法:

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    if ([segue.destinationViewController respondsToSelector:@selector(setMyData:)]) {
        [segue.destinationViewController performSelector:@selector(setMyData:) 
                                              withObject:myData];
    } 
}

所以只要你的目标视图控制器声明了一个公共属性，例如:

@property (nonatomic, strong) MyData *myData;

你可以在前面的视图控制器中设置这个属性。

在Swift 4.2中，我会这样做:

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    if let yourVC = segue.destination as? YourViewController {
        yourVC.yourData = self.someData
    }
}