Question

目标：具有内置分隔符的字符串应分为int和另一个字符串。如果分隔符序列'###'出现不止一次，则字符串应始终拼接在最后一个'###'。

是否有类似string.lastIndexOf（“###”）的运算符，就像在C＃中一样？

这就是我的解析器的样子：

func parseTuple(from string: String) -> (String, Int)? {
    let parsedString = string.components(separatedBy: "###")
    if let tupleString = String(parsedString[0]), let tupleInt = Int(parsedString[1]) {
        return (tupleString, tupleInt)
    } else {
        return nil
    }
}

Answer 1

String的{{3}}方法有.backwards个选项找到最后一次出现的字符串。然后substring(to:)和substring(from:)可以与。{1}}和func parseTuple(from string: String) -> (String, Int)? { if let theRange = string.range(of: "###", options: .backwards), let i = Int(string.substring(from: theRange.upperBound)) { return (string.substring(to: theRange.lowerBound), i) } else { return nil } }一起使用该范围的下限/上限以提取字符串的部分在分隔符之前/之后：

if let tuple = parseTuple(from: "Connect###Four###Player###7") {
    print(tuple)
    // ("Connect###Four###Player", 7)
}

示例：

func parseTuple(from string: String) -> (String, Int)? {

    if let theRange = string.range(of: "###", options: .backwards),
        let i = Int(string[theRange.upperBound...]) {
        return (String(string[...theRange.lowerBound]), i)
    } else {
        return nil
    }
}

Swift 4更新：

@Document
public class Artikal {

  public enum StatusZapisa {AKTIVAN, ARHIVIRAN}

  @Id
  private int sifra;
  private String naziv;
  @DBRef
  private KategorijaArtikla kategorija;
  private float cena;
  private int brojnoStanje;
  private Date datumKreiranjaZapisa;
  private boolean daLiPopunitiZalihe;
  private StatusZapisa statusZapisa;
  private int minimalnoStanje;

Answer 2

let input = "Connect###Four###Player###7"

let seperator = "###"
// split components
var components = input.components(separatedBy: seperator)
// remove the last component ...
components = Array(components.dropLast())
// ... and re-join the remaining ones
let output = components.joined(separator: seperator)

print(output)

打印：

Connect###Four###Player

如何在序列的最后一次出现时拆分字符串

2 个答案: