Question

我正在尝试设置一个根据特定条件显示不同结果的选择器。唯一的问题是我需要将dataSource（将显示的选择器数据）放在viewDidLoad之外。如果条件满足，我需要viewDidLoad加载适当的数据。但是，使用代码处理我所见过的拾取器，似乎dataSource数组放在viewDidLoad之外（我认为委托方法可以使用该值）。我只是想知道我是否可以将此dataSource放入viewDidLoad以便它能够识别它，或者是否有办法在viewDidLoad之外运行Switch或If语句类。

import UIKit
import Parse

class VC: UIViewController, UIPickerViewDataSource, UIPickerViewDelegate {

    var pickerData = [String]()

    @IBOutlet weak var namePicker: UIPickerView!

    override func viewDidLoad() {
        super.viewDidLoad()

        namePicker.dataSource = self
        namePicker.delegate = self


        //here I want the picker to show something different depending on the value of nameOfPerson. I changed the order of the array because the picker selects the first name of the array as the default. 
        switch nameOfPerson
        {
        case "Carly": 
            let pickerData = ["Carly","Jeff","Bob","Anita"]
        case "Jeff": 
            let pickerData = ["Jeff","Carly","Bob","Anita"]
        case "Bob": 
            let pickerData = ["Bob","Carly","Jeff","Anita"]
        case "Anita": 
            let pickerData = ["Anita","Carly","Jeff","Bob"]
        }
    }
   func numberOfComponents(in pickerView: UIPickerView) -> Int {
    return 1
    }
    func pickerView(_ pickerView: UIPickerView, numberOfRowsInComponent component: Int) -> Int {
    return pickerData.count
    }
    func pickerView(_ pickerView: UIPickerView, titleForRow row: Int, forComponent component: Int) -> String? {
    return pickerData[row]
    }
    func pickerView(_ pickerView: UIPickerView, didSelectRow row: Int, inComponent component: Int)
    {

    }
}

Answer 1

我只是想知道我是否可以将此dataSource放入viewDidLoad以便它能识别它，或者是否有办法在viewDidLoad和类中运行Switch或If语句

没有。数据源方法必须是方法，即在类的顶级函数，以便Objective-C查看它们。因此，数据本身必须对数据源方法可见，并且在viewDidLoad和数据源方法之间共享数据的唯一一致方式是作为属性，如var pickerData。

代码的错误是let：

    case "Carly": 
        let pickerData = ["Carly","Jeff","Bob","Anita"]
    case "Jeff": 
        let pickerData = ["Jeff","Carly","Bob","Anita"]
    case "Bob": 
        let pickerData = ["Bob","Carly","Jeff","Anita"]
    case "Anita": 
        let pickerData = ["Anita","Carly","Jeff","Bob"]
    }

从每个案例中删除let一词！否则，您只是设置一个临时局部变量pickerData，这不是您想要做的。您想要设置您声明的属性：

var pickerData = [String]()

....你的代码不是那样做的！你想要的是这个：

case "Carly": 
    self.pickerData = ["Carly","Jeff","Bob","Anita"]
case "Jeff": 
    self.pickerData = ["Jeff","Carly","Bob","Anita"]
case "Bob": 
    self.pickerData = ["Bob","Carly","Jeff","Anita"]
case "Anita": 
    self.pickerData = ["Anita","Carly","Jeff","Bob"]
}

（当然，你还需要default个案例。）

然后，最后，为了安全起见，重新加载数据：

self.namePicker.reloadAllComponents()

Answer 2

数据源需要在视图加载时设置，但您可以将其加载到您需要的任何位置。

内部viewDidLoad

picker.datasource = self

任何地方，您都可以设置实际数据;

picker.dataArray = "example, anything"

我不确定我完全理解你。如果你没有找到这个，请告诉我。

Xcode Swift 3 Picker：将DataSource放在ViewDidLoad中

2 个答案: