如何优化Swift中的getRange()函数以支持非连续日期场景？

首先需要给Days枚举添加顺序标识，这样才能判断日期是否连续。我们可以给它添加Int类型的rawValue并遵循CaseIterable协议：

enum Days: Int, CaseIterable, CustomStringConvertible {
    case Monday = 1, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday
    
    // 自定义字符串输出，避免默认枚举名的格式问题
    var description: String {
        switch self {
        case .Monday: return "Monday"
        case .Tuesday: return "Tuesday"
        case .Wednesday: return "Wednesday"
        case .Thursday: return "Thursday"
        case .Friday: return "Friday"
        case .Saturday: return "Saturday"
        case .Sunday: return "Sunday"
        }
    }
}

接下来重写getRange()相关函数，核心逻辑是将非连续的日期数组分割成连续区间，再根据需求生成不同格式的输出：

实现方案1：完整输出所有连续区间 #

适用于需要展示全部日期范围的场景，比如[.Monday, .Tuesday, .Saturday, .Sunday]输出Monday to Tuesday and Saturday to Sunday，[.Monday, .Saturday, .Sunday]输出Monday and Saturday to Sunday。

struct ChildView: View {
    let car: Car

    var body: some View {
        VStack {
            Text(getFullRangeString())
        }
    }

    private func getFullRangeString() -> String {
        guard !car.rent.isEmpty else { return "Empty" }
        
        // 1. 将日期数组按连续区间分组
        var ranges: [[Days]] = []
        var currentRange: [Days] = [car.rent[0]]
        
        for day in car.rent.dropFirst() {
            if day.rawValue == currentRange.last!.rawValue + 1 {
                currentRange.append(day)
            } else {
                ranges.append(currentRange)
                currentRange = [day]
            }
        }
        ranges.append(currentRange)
        
        // 2. 将每个区间转换为字符串
        let rangeStrings = ranges.map { range in
            range.count == 1 ? range[0].description : "\(range.first!.description) to \(range.last!.description)"
        }
        
        // 3. 拼接字符串，最后一个区间用"and"连接
        switch rangeStrings.count {
        case 0: return "Empty"
        case 1: return rangeStrings[0]
        default:
            let allButLast = rangeStrings.dropLast().joined(separator: ", ")
            return "\(allButLast) and \(rangeStrings.last!)"
        }
    }
}

实现方案2：简化输出（显示首个区间+“and others”） #

适用于只需要展示主要区间、其余简化的场景，比如[.Monday, .Tuesday, .Saturday, .Sunday]输出Monday to Tuesday and others，[.Monday, .Saturday, .Sunday]输出Monday and others。

private func getSimplifiedRangeString() -> String {
    guard !car.rent.isEmpty else { return "Empty" }
    
    // 生成首个区间的字符串
    var firstRange: [Days] = [car.rent[0]]
    for day in car.rent.dropFirst() {
        if day.rawValue == firstRange.last!.rawValue + 1 {
            firstRange.append(day)
        } else {
            break
        }
    }
    let firstRangeString = firstRange.count == 1 ? firstRange[0].description : "\(firstRange.first!.description) to \(firstRange.last!.description)"
    
    // 判断是否有其他非连续区间
    let hasOtherRanges = car.rent.count > firstRange.count
    return hasOtherRanges ? "\(firstRangeString) and others" : firstRangeString
}

实现方案3：仅显示首个连续区间 #

适用于只关注首个区间的场景，比如[.Monday, .Tuesday, .Saturday, .Sunday]输出Monday to Tuesday，[.Monday, .Saturday, .Sunday]输出Monday。

private func getFirstRangeString() -> String {
    guard !car.rent.isEmpty else { return "Empty" }
    
    var firstRange: [Days] = [car.rent[0]]
    for day in car.rent.dropFirst() {
        if day.rawValue == firstRange.last!.rawValue + 1 {
            firstRange.append(day)
        } else {
            break
        }
    }
    return firstRange.count == 1 ? firstRange[0].description : "\(firstRange.first!.description) to \(firstRange.last!.description)"
}

测试示例 #

// 测试场景1：非连续双区间
let car1 = Car(rent: [.Monday, .Tuesday, .Saturday, .Sunday])
// getFullRangeString() → "Monday to Tuesday and Saturday to Sunday"
// getSimplifiedRangeString() → "Monday to Tuesday and others"
// getFirstRangeString() → "Monday to Tuesday"

// 测试场景2：单个零散日期+连续区间
let car2 = Car(rent: [.Monday, .Saturday, .Sunday])
// getFullRangeString() → "Monday and Saturday to Sunday"
// getSimplifiedRangeString() → "Monday and others"
// getFirstRangeString() → "Monday"

// 测试场景3：空数组
let car3 = Car(rent: [])
// 所有函数返回 "Empty"

// 测试场景4：单个日期
let car4 = Car(rent: [.Wednesday])
// 所有函数返回 "Wednesday"

以上方案均无需修改现有Car结构体，直接通过优化函数即可适配所有场景，你可以根据业务需求选择对应的实现方式。

内容的提问来源于stack exchange，提问作者Alexnnd