问题描述

我正在尝试编写一个函数来将成千上万的数字呈现为 K 和 M例如:

I'm trying to write a function to present thousands and millions into K's and M's For instance:

1000 = 1k
1100 = 1.1k
15000 = 15k
115000 = 115k
1000000 = 1m

这是我到目前为止的地方:

Here is where I got so far:

func formatPoints(num: Int) -> String {
    let newNum = String(num / 1000)
    var newNumString = "(num)"
    if num > 1000 && num < 1000000 {
        newNumString = "(newNum)k"
    } else if num > 1000000 {
        newNumString = "(newNum)m"
    }

    return newNumString
}

formatPoints(51100) // THIS RETURNS 51K instead of 51.1K

如何让这个功能工作，我错过了什么?

How do I get this function to work, what am I missing?

推荐答案

func formatPoints(num: Double) ->String{
    let thousandNum = num/1000
    let millionNum = num/1000000
    if num >= 1000 && num < 1000000{
        if(floor(thousandNum) == thousandNum){
            return("(Int(thousandNum))k")
        }
        return("(thousandNum.roundToPlaces(1))k")
    }
    if num > 1000000{
        if(floor(millionNum) == millionNum){
            return("(Int(thousandNum))k")
        }
        return ("(millionNum.roundToPlaces(1))M")
    }
    else{
        if(floor(num) == num){
            return ("(Int(num))")
        }
        return ("(num)")
    }

}

extension Double {
    /// Rounds the double to decimal places value
    func roundToPlaces(places:Int) -> Double {
        let divisor = pow(10.0, Double(places))
        return round(self * divisor) / divisor
    }
}

如果数字是整数，更新后的代码现在不应返回 .0.例如，现在应该输出 1k 而不是 1.0k.我只是检查了 double 和它的 floor 是否相同.

The updated code should now not return a .0 if the number is whole. Should now output 1k instead of 1.0k for example. I just checked essentially if double and its floor were the same.

我在这个问题中找到了双重扩展名:将一个双精度值舍入到 x 个swift中的小数位

I found the double extension in this question: Rounding a double value to x number of decimal places in swift

这篇关于Swift 2.0 将 1000 格式化为友好的 K的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！