填充快速字符串以进行打印

Question

我正在尝试打印所有填充到相同宽度的字符串列表。

在 C 语言中，我会使用 printf("%40s", cstr), 之类的东西，其中 cstr 是 C 字符串。

在 Swift 中，我能想到的最好的方法是：

line += String(format: "%40s",string.cStringUsingEncoding(<someEncoding>))

有没有更好的办法？

Answer 1

在 Swift 3 中，您可以使用：

let str = "Test string"
let paddedStr = str.padding(toLength: 20, withPad: " ", startingAt: 0)

结果字符串："Test string "

如果您需要向左填充文本（右对齐），您可以编写以下函数作为String的扩展：

extension String {
    func leftPadding(toLength: Int, withPad character: Character) -> String {
        let newLength = self.characters.count
        if newLength < toLength {
            return String(repeatElement(character, count: toLength - newLength)) + self
        } else {
            return self.substring(from: index(self.startIndex, offsetBy: newLength - toLength))
        }
    }
}

所以如果你写：

let str = "Test string"
let paddedStr = str.leftPadding(toLength: 20, withPad: " ")

结果字符串：" Test string"

在 Swift 4.1 中，substring 方法已被弃用，并且有许多新方法可以获取子字符串。 prefix、suffix 或使用Range<String.Index> 为String 下标。

对于之前的扩展，我们可以使用suffix 方法来实现相同的结果。由于suffix方法返回的是String.SubSequence，所以在返回之前需要转换成String。

extension String {
    func leftPadding(toLength: Int, withPad character: Character) -> String {
        let stringLength = self.count
        if stringLength < toLength {
            return String(repeatElement(character, count: toLength - stringLength)) + self
        } else {
            return String(self.suffix(toLength))
        }
    }
}

Answer 2

对于 Swift >= 3

line += string.padding(toLength: 40, withPad: " ", startingAt: 0)

对于斯威夫特

NSString 有 stringByPaddingToLength: 方法：

line += string.stringByPaddingToLength(40, withString: " ", startingAtIndex: 0)

Answer 3

extension RangeReplaceableCollection where Self: StringProtocol {
    func paddingToLeft(upTo length: Int, using element: Element = " ") -> SubSequence {
        return repeatElement(element, count: Swift.max(0, length-count)) + suffix(Swift.max(count, count-length))
    }
}

---

"123".paddingToLeft(upTo: 5)              //  "  123"
"123".paddingToLeft(upTo: 5, using: "0")  //  "00123"
"123".paddingToLeft(upTo: 3, using: "0")  //    "123"
"$199.99".dropLast(3).paddingToLeft(upTo: 10, using: "_")  //  "______$199"

要复制与 padding(toLength:, withPad:, startingAt:) 相同的行为，我们可以将 rotateTo left 功能添加到 RangeReplaceableCollection

extension RangeReplaceableCollection {
    func rotatingLeft(positions: Int) -> SubSequence {
        let index = self.index(startIndex, offsetBy: positions, limitedBy: endIndex) ?? endIndex
        return self[index...] + self[..<index]
    }
}

并按如下方式实现：

extension RangeReplaceableCollection where Self: StringProtocol {
    func paddingToLeft<S: StringProtocol & RangeReplaceableCollection>(upTo length: Int, with string: S, startingAt index: Int = 0) -> SubSequence {
        let string = string.rotatingLeft(positions: index)
        return repeatElement(string, count: length-count/string.count)
            .joined().prefix(length-count) + suffix(Swift.max(count, count-length))
    }
}

---

"123".paddingToLeft(upTo: 10, with: "abc", startingAt: 2)   //  "cabcabc123"
"123".padding(toLength: 10, withPad: "abc", startingAt: 2)  //  "123cabcabc"

Answer 4

将所有字符串格式代码放入extension 并在任何你想要的地方重复使用。

extension String {
    func padding(length: Int) -> String {
        return self.stringByPaddingToLength(length, withString: " ", startingAtIndex: 0)
    }

    func padding(length: Int, paddingString: String) -> String {
        return self.stringByPaddingToLength(length, withString: paddingString, startingAtIndex: 0)
    }
}

var str = "str"
print(str.padding(10)) // "str       "
print(str.padding(10, paddingString: "+")) // "str+++++++"

Answer 5

以下两个函数返回一个填充到给定宽度的字符串，左对齐或右对齐。它是纯 Swift 4，没有 NSString，也没有 C 字符串。您可以选择是否截断长于填充宽度的字符串。

extension String {
    func rightJustified(width: Int, truncate: Bool = false) -> String {
        guard width > count else {
            return truncate ? String(suffix(width)) : self
        }
        return String(repeating: " ", count: width - count) + self
    }

    func leftJustified(width: Int, truncate: Bool = false) -> String {
        guard width > count else {
            return truncate ? String(prefix(width)) : self
        }
        return self + String(repeating: " ", count: width - count)
    }
}

Answer 6

这是我的解决方案，专门针对 String，但我确信比我更聪明的人可以使它更通用。

extension String {
    func frontPadding(toLength length: Int, withPad pad: String, startingAt index: Int) -> String {
        return String(String(self.reversed()).padding(toLength: length, withPad: pad, startingAt: index).reversed())
    }
}

Answer 7

对于左填充，您可以使用双反转技巧：

String(String(s.reversed()).padding(toLength: 5, withPad: "0", startingAt: 0).reversed())

当然，你可以把它包装成一个扩展：

extension String {
    func leftPadding(toLength: Int, withPad: String) -> String {
        String(String(reversed()).padding(toLength: toLength, withPad: withPad, startingAt: 0).reversed())
    }
}

Answer 8

import Foundation   // for NSString.padding()

/**
 *  Custom Extension's API
 *  ------------------------------
 *  • str.padEnd(_:_:)
 *  • str.padStart(_:_:)
 *  ------------------------------
 *  • int.padStart(_:_:forceSign:)
 */

extension String {

    // str.padEnd(8, "-")
    func padEnd(_ length: Int, _ pad: String) -> String {
        return padding(toLength: length, withPad: pad, startingAt: 0)
    }

    // str.padStart(8, "*")
    func padStart(_ length: Int, _ pad: String) -> String {
        let str = String(self.reversed())
        return String(str.padEnd(length, pad).reversed())
    }
}

extension Int {

    // int.padStart(8)
    func padStart(

        _  length: Int,             // total length
        _     pad: String = "0",    // pad character
        forceSign: Bool   = false   // force + sign if positive

    ) -> String {

        let isNegative = self < 0
        let n = abs(self)
        let str = String(n).padStart(length, pad)

        return 
            isNegative ? "- " + str :
            forceSign  ? "+ " + str :
            str
    }
}

// test run
let s1 = "abc"

[
    s1.padEnd(15, "*"),                 // abc************
    s1.padStart(15, "*"),               // ************abc
    3.padStart(8, forceSign: true),     // + 00000003
    (-125).padStart(8)                  // - 00000125

].forEach{ print($0) }