PickerView 带小数点答案

【问题标题】：PickerView with decimal pointPickerView 带小数点
【发布时间】：2018-03-08 03:44:57
【问题描述】：

我使用具有四个组件的 UIPickerView 来输入数字而无需键盘。我添加了一个小数点，只有一个“。”在 UILabelView 中。查看屏幕以了解想法。

我的 UIPickerView 代码是：

func numberOfComponents(in weightPickerView: UIPickerView) -> Int {
    return 4
}

func pickerView(_ pickerView: UIPickerView, numberOfRowsInComponent component: Int) -> Int {
    return loopingMargin * numbers.count
}

func pickerView(_ pickerView: UIPickerView, titleForRow row: Int, forComponent component: Int) -> String? {
  return numbers[row % numbers.count]
}

func pickerView(_ pickerView: UIPickerView, didSelectRow row: Int, inComponent component: Int) {
    let currentIndex = row % numbers.count
    digits[component] = currentIndex
    let weightString = "\(digits[0])\(digits[1]).\(digits[2])\(digits[3])"
    weightField.text = weightString
}

在 viewDidLoad() 里面的部分代码是：

override func viewDidLoad() {
    super.viewDidLoad()        
    // Do any additional setup after loading the view.

    weightPickerView.dataSource = self
    weightPickerView.delegate = self
    weightPickerView.selectRow((loopingMargin / 2) * numbers.count, inComponent: 0, animated: false)
    weightPickerView.selectRow((loopingMargin / 2) * numbers.count, inComponent: 1, animated: false)
    weightPickerView.selectRow((loopingMargin / 2) * numbers.count, inComponent: 2, animated: false)
    weightPickerView.selectRow((loopingMargin / 2) * numbers.count, inComponent: 3, animated: false)

}

我的问题...我想添加第五个组件，只有一个“。”在其中（因此空间分布均匀）。我该怎么做？

【问题讨论】：

您已经知道如何添加 4 个组件。添加第 5 个有什么问题？
添加一个 '.'在里面...

标签： swift uipickerview

【解决方案1】：

您只需要更新您的选择器视图方法以提供额外的组件。假设你想要它在中间，你的代码需要是：

func numberOfComponents(in weightPickerView: UIPickerView) -> Int {
    return 5
}

func pickerView(_ pickerView: UIPickerView, numberOfRowsInComponent component: Int) -> Int {
    return component == 2 ? 1 : loopingMargin * numbers.count
}

func pickerView(_ pickerView: UIPickerView, titleForRow row: Int, forComponent component: Int) -> String? {
  return component == 2 ? "." : numbers[row % numbers.count]
}

func pickerView(_ pickerView: UIPickerView, didSelectRow row: Int, inComponent component: Int) {
    if component != 2 {
        var index = component
        if component > 2 {
            index -= 1
        }
        let currentIndex = row % numbers.count
        digits[index] = currentIndex
        let weightString = "\(digits[0])\(digits[1]).\(digits[2])\(digits[3])"
        weightField.text = weightString
    }
}

override func viewDidLoad() {
    super.viewDidLoad()        

    weightPickerView.dataSource = self
    weightPickerView.delegate = self
    weightPickerView.selectRow((loopingMargin / 2) * numbers.count, inComponent: 0, animated: false)
    weightPickerView.selectRow((loopingMargin / 2) * numbers.count, inComponent: 1, animated: false)
    weightPickerView.selectRow((loopingMargin / 2) * numbers.count, inComponent: 3, animated: false)
    weightPickerView.selectRow((loopingMargin / 2) * numbers.count, inComponent: 4, animated: false)
}

假设. 代表小数分隔符，请记住，世界各地的许多用户不使用. 作为小数分隔符。所以你真的应该表现出适当的性格。您可以获得NumberFormatter 的decimalSeparator 属性。

【讨论】：

【解决方案2】：

你可以试试

func numberOfComponents(in weightPickerView: UIPickerView) -> Int {
    return 5
}


 func pickerView(_ pickerView: UIPickerView, titleForRow row: Int, forComponent component: Int) -> String? {
   if(component == 2)
   {
      return "."
   }
   else
   {
       return numbers[row % numbers.count]
   }
}

【讨论】：