ios - 快速将UIPickerView设置为UITextField的输入时出错

我是快速编程的新手，我想将 UIPickerView 设置为 UITextField 的输入。这是我的代码:

import UIKit

class ViewController: UIViewController, UIPickerViewDataSource, UIPickerViewDelegate {

    @IBOutlet var  selectTransportCompany: UITextField!
    var data = ["DHL Express", "Fed Ex", "TNT", "Express Mail"]
    var picker = UIPickerView()

    override func viewDidLoad() {
        super.viewDidLoad()
        picker.delegate = self
        picker.dataSource = self
        selectTransportCompany.inputView = picker
    }

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

    func pickerView(pickerView: UIPickerView, numberOfRowsInComponent component: Int) -> Int{
        return data.count
    }

    func pickerView(pickerView: UIPickerView, didSelectRow row: Int, inComponent component: Int) {
        selectTransportCompany.text = data[row]

    }

    func pickerView(pickerView: UIPickerView, titleForRow row: Int, forComponent component: Int) -> String? {
        return data[row]
    }


}

但是，当我运行此代码时，出现错误提示

线程1:EXC_BAD_INSTRUCTION

进入调试器，我得到以下信息:

致命错误:解开Optional值时意外发现nil

有人知道这个问题是什么，我也关注this tutorial。

最佳答案

您只需要将视图控制器文本字段出口连接到故事板selectTransportCompany textField。

关于ios - 快速将UIPickerView设置为UITextField的输入时出错，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/34367864/