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swift - 在每个循环中快速更改不同的标签 labelData

问题描述

所以我想将这段代码简化为一个简单的 for 循环。但是 UITextFields 和 UILabels 给我带来了一些困难。

    @IBOutlet var thingsIUse: UITextField!
    @IBOutlet var orThis: UILabel!

    saveProgress(year: year, season: season, day: "1", value: "\(Int(day1new.text!)! + Int(day1newExtra.text!)!)")
    saveProgress(year: year, season: season, day: "2", value: "\(Int(day2new.text!)! + Int(day2newExtra.text!)!)")
    saveProgress(year: year, season: season, day: "3", value: "\(Int(day3new.text!)! + Int(day3newExtra.text!)!)")
    saveProgress(year: year, season: season, day: "4", value: "\(Int(day4new.text!)! + Int(day4newExtra.text!)!)")
    saveProgress(year: year, season: season, day: "5", value: "\(Int(day5new.text!)! + Int(day5newExtra.text!)!)")
    saveProgress(year: year, season: season, day: "6", value: "\(Int(day6new.text!)! + Int(day6newExtra.text!)!)")
    saveProgress(year: year, season: season, day: "1Taki", value: day1new.text!)
    saveProgress(year: year, season: season, day: "2Taki", value: day2new.text!)
    saveProgress(year: year, season: season, day: "3Taki", value: day3new.text!)
    saveProgress(year: year, season: season, day: "4Taki", value: day4new.text!)
    saveProgress(year: year, season: season, day: "5Taki", value: day5new.text!)
    saveProgress(year: year, season: season, day: "6Taki", value: day6new.text!)

对于这样的事情...

    for k in 1...10{
        saveProgress(year: year, season: season, day: "\(k)", value: "\(Int(day\(k)new.text!)! + Int(day\(k)newExtra.text!)!)")
        saveProgress(year: year, season: season, day: "\(k)Taki", value: day\(k)new.text!)
    }

有什么建议么？

标签： swifttagsuitextfielduilabelsimplify

解决方案

你可以通过几种方式解决这个问题。一种是使用 IBOutlet 集合。迅速地，您将它们定义为：

@IBOutlet var dayLabels: [UILabel]!

然后确保进入 Xib/Storyboard 并通过单击收集名称旁边的小“+”并将其拖动到各种标签来添加对标签的引用。由于这是一个集合，您可以像这样拖动到许多标签上，只需确保按照您希望它们更新的顺序添加它们。

解决此问题的另一种方法是创建要在运行时动态更新的标签数组，如下所示：

let labelsToUpdate: [UILabel] = [day1new,day2new,day3new,day4new,day5new,day6new]
for k in 0...5 {
    let day = "\(k+1)"
    let label = labelsToUpdate[k]
    ...
}

你可以做一些其他的优化来保持你的代码干净，但这将是实现你想要的一个好的开始。

祝你好运！

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