这是一个通用的方法，它允许您在指定的日期中添加/减去任何类型的单位(年/月/日/小时/秒等)。

使用Swift 2.2

func addUnitToDate(unitType: NSCalendarUnit, number: Int, date:NSDate) -> NSDate {

    return NSCalendar.currentCalendar().dateByAddingUnit(
        unitType,
        value: number,
        toDate: date,
        options: NSCalendarOptions(rawValue: 0))!

}

print( addUnitToDate(.Day, number: 1, date: NSDate()) ) // Adds 1 Day To Current Date
print( addUnitToDate(.Hour, number: 1, date: NSDate()) ) // Adds 1 Hour To Current Date
print( addUnitToDate(.Minute, number: 1, date: NSDate()) ) // Adds 1 Minute To Current Date

// NOTE: You can use negative values to get backward values too

只是为了好玩，通过一些扩展和操作符重载，你可以得到一些不错的东西，比如:

let today = Date()
let tomorrow = today + 1.days

, or

var date = Date()
date += 1.months

下面是支持代码:

extension Calendar {
    struct ComponentWithValue {
        let component: Component
        let value: Int
    }
}

extension Int {
    var days: Calendar.ComponentWithValue {
        .init(component: .day, value: self)
    }
    
    var months: Calendar.ComponentWithValue {
        .init(component: .month, value: self)
    }
}

func +(_ date: Date, _ amount: Calendar.ComponentWithValue) -> Date {
    Calendar.current.date(byAdding: amount.component, value: amount.value, to: date)!
}

func +(_ amount: Calendar.ComponentWithValue, _ date: Date) -> Date {
    date + amount
}

func +=(_ date: inout Date, _ amount: Calendar.ComponentWithValue) {
    date = date + amount
}

代码是最少的，并且可以很容易地扩展到允许.月，.年，.小时等。还可以无缝添加对减法(-)的支持。

虽然在+操作符的实现中有一个强制的展开，但是不确定在哪种情况下日历可以返回nil日期。

斯威夫特5

let today = Date()
let nextDate = Calendar.current.date(byAdding: .day, value: 1, to: today)

objective - c

NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
// now build a NSDate object for the next day
NSDateComponents *offsetComponents = [[NSDateComponents alloc] init];
[offsetComponents setDay:1];
NSDate *nextDate = [gregorian dateByAddingComponents:offsetComponents toDate: [NSDate date] options:0];

Swift 4更新:

let now = Date() // the current date/time
let oneDayFromNow = Calendar.current.date(byAdding: .day, value: 1, to: now) // Tomorrow with same time of day as now

字符串扩展:转换String_Date >日期

extension String{
  func DateConvert(oldFormat:String)->Date{ // format example: yyyy-MM-dd HH:mm:ss 
    let isoDate = self
    let dateFormatter = DateFormatter()
    dateFormatter.locale = Locale(identifier: "en_US_POSIX") // set locale to reliable US_POSIX
    dateFormatter.dateFormat = oldFormat
    return dateFormatter.date(from:isoDate)!
  }
}

日期扩展:转换日期>字符串

extension Date{
 func DateConvert(_ newFormat:String)-> String{
    let formatter = DateFormatter()
    formatter.dateFormat = newFormat
    return formatter.string(from: self)
 }
}

日期扩展:获取+/-日期

extension String{
  func next(day:Int)->Date{
    var dayComponent    = DateComponents()
    dayComponent.day    = day
    let theCalendar     = Calendar.current
    let nextDate        = theCalendar.date(byAdding: dayComponent, to: Date())
    return nextDate!
  }

 func past(day:Int)->Date{
    var pastCount = day
    if(pastCount>0){
        pastCount = day * -1
    }
    var dayComponent    = DateComponents()
    dayComponent.day    = pastCount
    let theCalendar     = Calendar.current
    let nextDate        = theCalendar.date(byAdding: dayComponent, to: Date())
    return nextDate!
 }
}

用法:

let today = Date()
let todayString = "2020-02-02 23:00:00"
let newDate = today.DateConvert("yyyy-MM-dd HH:mm:ss") //2020-02-02 23:00:00
let newToday = todayString.DateConvert(oldFormat: "yyyy-MM-dd HH:mm:ss")//2020-02-02
let newDatePlus = today.next(day: 1)//2020-02-03 23:00:00
let newDateMinus = today.past(day: 1)//2020-02-01 23:00:00

参考:来自多重问题 我如何添加1天到一个NSDate? 数学函数转换正整数为负和负到正? 将NSString转换为NSDate(然后再转换回来)

iOS 8+， OSX 10.9+， Objective-C

NSCalendar *cal = [NSCalendar currentCalendar];    
NSDate *tomorrow = [cal dateByAddingUnit:NSCalendarUnitDay 
                                   value:1 
                                  toDate:[NSDate date] 
                                 options:0];