作者:白羊幸福的佳佳 | 来源:互联网 | 2023-01-31 15:25
Iamgettingnumberofdays,Startdateandenddatefromaservice.Iwanttogetthelistofallda
I am getting number of days,Start date and end date from a service. I want to get the list of all dates in between start date and end date. Let's say my start date is 2017/08/15 and end date is 2017/08/16 and number of days is 2.
我从服务中获取天数,开始日期和结束日期。我想获得开始日期和结束日期之间所有日期的列表。假设我的开始日期是2017/08/15,结束日期是2017/08/16,天数是2。
But I am getting the date list like this.
但我得到这样的日期列表。
##### CONVERTED STRING DATE 2017-08-16
##### CONVERTED STRING DATE 2017-08-17
##### CONVERTED STRING DATE 2017-08-18
And I have another date like this
我有另一个这样的日期
Start date 2017/08/23 end date 2017/09/01 and number of days 8. then I get the list like this.
开始日期2017/08/23结束日期2017/09/01和天数8.然后我得到这样的列表。
##### CONVERTED STRING DATE 2017-08-24
##### CONVERTED STRING DATE 2017-08-25
##### CONVERTED STRING DATE 2017-08-28
##### CONVERTED STRING DATE 2017-08-29
##### CONVERTED STRING DATE 2017-08-30
##### CONVERTED STRING DATE 2017-08-31
##### CONVERTED STRING DATE 2017-09-01
This is how I get the dates array
这就是我获取日期数组的方法
numberOfDates=Int(ceil(numOfDay))
//numberOfDates=numberOfDates-1
let arrayDates=self.generateDates(startDate: startDate, addbyUnit: .day, value: numberOfDates)
This is how my date calculation method
这是我的日期计算方法
internal func generateDates(startDate :Date?, addbyUnit:Calendar.Component, value : Int) -> [Date]
{
//print("####START DATE#######\(startDate)")
var calendar = Calendar.current
calendar.timeZOne=TimeZone.current
var datesArray: [Date] = [Date] ()
for i in 0 ... value {
var addAmount:Int!
if(value==0)
{
addAmount=0
}
else
{
addAmount=1
}
if let newDate = calendar.date(byAdding: addbyUnit, value: i + addAmount, to: startDate!) {
let strDayName=self.getDayName(mydate: newDate)
if (strDayName != "Saturday" && strDayName != "Sunday")
{
datesArray.append(newDate)
}
}
}
return datesArray
}
My problem is sometimes the date list is wrong (1st scenario) but its correct in the 2nd scenario.
我的问题有时是日期列表错误(第一种情况),但在第二种情况下它是正确的。
2 个解决方案