设置UIPickerView的默认值：索引超出范围

Posted 2023-02-24

我正在尝试为PickerView元素设置默认值，但是运行时出现错误：索引超出了键，我的意图是从“ Difficult”选项开始，默认为nvel = 3。

@IBOutlet weak var spinner: UIPickerView!

override func viewDidLoad() 
    super.viewDidLoad()
    self.spinner.delegate = self
    self.spinner.dataSource = self
    nivel = 3
    lista_parametros = ["All","Easy","Medium","Dificult"]
    self.spinner.selectRow(4, inComponent: nivel, animated: true) /// the error is here !! when running 


func numberOfComponents(in pickerView: UIPickerView) -> Int 
    return 1


func pickerView(_ pickerView: UIPickerView, numberOfRowsInComponent component: Int) -> Int 
    return lista_parametros.count


func pickerView(_ pickerView: UIPickerView, titleForRow row: Int, forComponent component: Int) -> String? 
     return lista_parametros[row]
 

 func pickerView(_ pickerView: UIPickerView, didSelectRow row: Int, inComponent component: Int) 
     nivel = row
     UserDefaults.standard.set(nivel,forKey: "nivel")
 

答案

索引从0开始。因此，您应该将其修改为：

另一答案

必须在第一个参数处进行选择：