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带属性字符串的UIPickerView

uipickerview swift ios

MacUserT · 技术社区 · 6 年前

为了简单起见,我有一个简单的程序,在picker视图的两个组件中显示两个数组的内容。程序只是一个视图控制器和一个选择器视图。代码呈现给她:

import UIKit

类ViewController:UIViewController、UIPickerViewDeleteGate、UIPickerViewDataSource{

var familyNames = [String]()
var fontName = ""
let firstArray = Array(0...99)
let secondArray = Array(0...99)
let fontCount = 0

@IBOutlet weak var samplePickerView: UIPickerView!
@IBOutlet weak var fontLabel: UILabel!

override func viewDidLoad() {
    super.viewDidLoad()

    familyNames = UIFont.familyNames.sorted()
    let fontNames = UIFont.fontNames(forFamilyName: familyNames[17])
    fontName = fontNames.first!

    samplePickerView.delegate = self
    samplePickerView.dataSource = self
}

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

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

    if component == 0 {
        return firstArray.count
    } else {
        return secondArray.count
    }
}

func pickerView(_ pickerView: UIPickerView, attributedTitleForRow row: Int, forComponent component: Int) -> NSAttributedString? {

    var rowTitle = ""
    let font = UIFont(name: fontName, size: 18.0)
    let stringDictionary = [NSAttributedString.Key.font: font]

    switch component {
    case 0:
        rowTitle = String(format: "%03d", firstArray[row])
    case 1:
        rowTitle = String(format: "%03d", secondArray[row])
    default:
        break
    }

    let returnString = NSAttributedString(string: rowTitle, attributes: stringDictionary as [NSAttributedString.Key : Any])
    print(returnString)
    return returnString
}

picker视图现在应该在Bradley Hand中显示标题,因此很容易发现它是有效的。不幸的是,picker视图没有在属性化字符串中显示标题。委托方法返回的字符串是一个属性化的字符串,因此应该可以工作。图片显示情况并非如此。我做错什么了?

2 回复 | 直到 6 年前

MacUserT 6 年前

在没有找到解决方案之后,我请求苹果公司的DTS来了解我做错了什么。显然,我没有做错什么,但是方法“pickerView(\pickerView:UIPickerView,attributeditleforrow row:Int,forComponent:Int)->nsattributestring?“不允许您对属性化字符串进行太多操作。苹果给出的答案是:“UIPickerViews attributeditleforrow delegate不会更改字体属性,但会对字体颜色等其他属性起作用。”

幸运的是,因为我想更改字体大小和字体,所以有一种方法可以在UIPickerView中获得不同的字体和字体大小。苹果提供的解决方案是:“如果你想使用nsattributestring中的所有字体属性,可以使用viewForRow并返回UILabel作为替代。”

现在代码如下所示: 设firstArray=Array(0…99) 设secondArray=Array(0…99)

@IBOutlet weak var samplePickerView: UIPickerView!
@IBOutlet weak var fontLabel: UILabel!

override func viewDidLoad() {
    super.viewDidLoad()

    familyNames = UIFont.familyNames.sorted()
    let fontNames = UIFont.fontNames(forFamilyName: familyNames[17])
    fontName = fontNames.first!

    samplePickerView.delegate = self
    samplePickerView.dataSource = self
}

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

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

    if component == 0 {
        return firstArray.count
    } else {
        return secondArray.count
    }
}

func pickerView(_ pickerView: UIPickerView, viewForRow row: Int, forComponent component: Int, reusing view: UIView?) -> UIView {

    var rowTitle = ""
    let pickerLabel = UILabel()

    pickerLabel.textColor = UIColor.blue

    switch component {
        case 0:
            rowTitle = String(format: "%03d", firstArray[row])
        case 1:
            rowTitle = String(format: "%03d", secondArray[row])
        default:
            break
    }

    pickerLabel.text = rowTitle
    pickerLabel.font = UIFont(name: fontName, size: 24.0)
    pickerLabel.textAlignment = .center

    return pickerLabel
}

Community T.Woody 4 年前

问题来了

familyNames = UIFont.familyNames.sorted()

字体族 ,但是 UIFont 初始值设定项作为一个参数,而不是它的 . 从 Apple Documentation :

家庭名称

字体系列名称与字体的基本名称相对应,例如Times New Roman。您可以将返回的字符串传递给 fontNames(forFamilyName:) 方法检索可用于该族的字体名称列表。 然后可以使用相应的字体名称来检索实际的字体对象。

UIFont.fontNames(forFamilyName: String)

另外,不要通过像这样的神奇索引号访问字体系列名称 familyNames[17] ,这是一个糟糕的方法。你可能会越界,这将使你的应用程序崩溃。

let pickerFontName = "MyAwesomeFont"

// later in your program
let font = UIFont(name: pickerFontName, size: 18.0)