基本上,就像标题说的那样。我想知道如何添加1天的NSDate。

如果它是:

21st February 2011

它将变成:

22nd February 2011

或者如果它是:

31st December 2011

它将变成:

1st January 2012.

当前回答

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。

其他回答

你可以使用NSDate的方法- (id)dateByAddingTimeInterval:(NSTimeInterval)秒,其中秒为60 * 60 * 24 = 86400

Swift 4,如果你真正需要的是24小时轮班(60*60*24秒)而不是“1个日历天”

未来: let dayAhead = Date(timeIntervalSinceNow: TimeInterval(86400.0))

过去: let dayAgo = Date(timeIntervalSinceNow: TimeInterval(-86400.0))

在Swift 2.1.1和xcode 7.1 OSX 10.10.5中,你可以使用函数添加任意数量的天数

func addDaystoGivenDate(baseDate:NSDate,NumberOfDaysToAdd:Int)->NSDate
{
    let dateComponents = NSDateComponents()
    let CurrentCalendar = NSCalendar.currentCalendar()
    let CalendarOption = NSCalendarOptions()

    dateComponents.day = NumberOfDaysToAdd

    let newDate = CurrentCalendar.dateByAddingComponents(dateComponents, toDate: baseDate, options: CalendarOption)
    return newDate!
}

函数调用,将当前日期增加9天

var newDate = addDaystoGivenDate(NSDate(), NumberOfDaysToAdd: 9)
print(newDate)

函数调用,将当前日期减少80天

newDate = addDaystoGivenDate(NSDate(), NumberOfDaysToAdd: -80)
 print(newDate)

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)

使用下面的函数,并使用days参数来获取日期daysAhead/daysBehind,只需将参数传递为正,表示未来日期,为负,表示之前日期:

+ (NSDate *) getDate:(NSDate *)fromDate daysAhead:(NSUInteger)days
{
    NSDateComponents *dateComponents = [[NSDateComponents alloc] init];
    dateComponents.day = days;
    NSCalendar *calendar = [NSCalendar currentCalendar];
    NSDate *previousDate = [calendar dateByAddingComponents:dateComponents
                                                     toDate:fromDate
                                                    options:0];
    [dateComponents release];
    return previousDate;
}