我想在我的应用程序中添加一个功能，用户必须从 UIPickerView 中选择一个类别，单击按钮后，程序将打开从 UIPickerView 中的类别中选择的视图。例如，如果 UIPickerView 上选择的选项之一是“关于页面”，我希望有一个程序可以在单击按钮时打开“关于页面”。
但不是状态 - 我会有类别，我不确定如何在 UIPickerView 中设置值并打开特定视图。
【问题讨论】：
标签： swift button uiviewcontroller uipickerview
实现UIPickerViewDelegate方法并给每个VC一个stroryboard Identifier
var lastId:String?
var ids = ["secondVC","thirdVC"] // picker data source and identifier names
func pickerView(_ pickerView: UIPickerView,
didSelectRow row: Int,
inComponent component: Int) {
lastId = ids[row]
}
@IBAction func goClicked(_ sender:UIButton) {
if let id = lastId {
let vc = storyboard.instantiateViewController(withIdentifier:id)!
self.present(vc, animated: true, completion: nil) // or push
}
else {
// select a vc alert
}
}
【讨论】：
performSegue(withIdentifier:sender:) 一起使用的 segue 标识符。
如果您想以编程方式创建选择器视图（我最推荐这种方式），此解决方案应该适合您：
class MyViewController : UIViewController, UIPickerViewDelegate, UIPickerViewDataSource {
let pickerView = UIPickerView()
let titles = ["view1", "view2"]
override func viewDidLoad() {
super.viewDidLoad()
view.backgroundColor = .white
pickerView.delegate = self
pickerView.dataSource = self
view.addSubview(pickerView)
}
func pickerView(_ pickerView: UIPickerView, numberOfRowsInComponent component: Int) -> Int {
return 2
}
func pickerView(_ pickerView: UIPickerView, titleForRow row: Int, forComponent component: Int) -> String? {
return titles[row]
}
func numberOfComponents(in pickerView: UIPickerView) -> Int {
return 1
}
func pickerView(_ pickerView: UIPickerView, didSelectRow row: Int, inComponent component: Int) {
if row == 0 {
let view1 = viewController1()
self.present(view1, animated: true, completion: nil)
}
if row == 1 {
let view2 = viewController2()
self.present(view2, animated: true, completion: nil)
}
}
}
class viewController1 : UIViewController {
override func viewDidLoad() {
super.viewDidLoad()
view.backgroundColor = UIColor.white
}
}
class viewController2 : UIViewController {
override func viewDidLoad() {
super.viewDidLoad()
view.backgroundColor = UIColor.white
}
}希望这对你也有用；）
【讨论】：