不符合协议UIPickerViewDataSource

Question

我不知道我的编码有什么问题.我试着按照教程,但同样的错误发生.

错误:

类型'FourthViewController'不符合协议'UIPickerViewDataSource'

这是我的代码:

import UIKit

let characters = ["Jaja Bink", "Luke", "Han Solo", "Princess Leia"];

let weapons = ["LightSaber", "Pistol", "Keris"];

class FourthViewController: UIViewController, UIPickerViewDataSource, UIPickerViewDelegate {

    @IBOutlet weak var doublePicker: UIPickerView!

    override func viewDidLoad() {
        super.viewDidLoad()

        // Do any additional setup after loading the view.
    }

    override func didReceiveMemoryWarning() {
        super.didReceiveMemoryWarning()
        // Dispose of any resources that can be recreated.
    }

    func numberOfComponentsInPickerView(pickerView: UIPickerView) -> Int {
        return 2
    }


    func pickerView(pickerView: UIPickerView,
                    titleForRow row: Int,
                                forComponent component: Int) -> String? {

        if component == 0 {
            return characters[row]
        } else {
            return weapons[row]
        }

    }

    func pickerView(pickerView: UIPickerView, numberOfRowsInComponent component: Int) -> Int? {
        if component == 0 {
            return characters.count
        } else {
            return weapons.count
        }
    }

}

Answer 1

更换协议方法的弃用版本UIPickerViewDataSource,如果您正在使用Swift 3可能适合您.

不推荐的协议方法

func numberOfComponentsInPickerView(pickerView: UIPickerView) -> Int {
    return 1
}

Swift中的最新协议方法3

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

注意:其他所需的协议方法也是如此.即

推荐使用:

func pickerView(pickerView: UIPickerView!, numberOfRowsInComponent component: Int) -> Int

最新版本:

func pickerView(_ pickerView: UIPickerView, numberOfRowsInComponent component: Int) -> Int

Answer 2

因为您所需的委托方法不正确。

将其替换为：

func pickerView(pickerView: UIPickerView, numberOfRowsInComponent component: Int) -> Int {

    if component == 0 {
        return characters.count
    } else {
        return weapons.count
    }
}

您的委托方法返回的Int?结果不正确。