Get the last day of the month in time. Time

Question

When I have time.Time :

 // January, 29th t, _ := time.Parse("2006-01-02", "2016-01-29") 

How can I get time.Time , which represents January 31st? This example is trivial, but when there is a date in February, the last day can be 28 or 29.

Answer 1

Packet time

func Date

 func Date(year int, month Month, day, hour, min, sec, nsec int, loc *Location) Time 

Date returns the time corresponding to

yyyy-mm-dd hh: mm: ss + nsec nanoseconds

in the corresponding zone during this time in this place.

The values ​​of month, day, hour, min, sec and nsec can be outside their usual ranges and will be normalized during conversion. For example, October 32 is converted to November 1.

For example, normalizing a date,

 package main import ( "fmt" "time" ) func main() { // January, 29th t, _ := time.Parse("2006-01-02", "2016-01-29") fmt.Println(t.Date()) // January, 31st y,m,_ := t.Date() lastday:= time.Date(y,m+1,0,0,0,0,0,time.UTC) fmt.Println(lastday.Date()) } 

Output:

 2016 January 29 2016 January 31 

Answer 2

You can write the function yourself, perhaps something like this:

 func daysInMonth(month, year int) int { switch time.Month(month) { case time.April, time.June, time.September, time.November: return 30 case time.February: if year%4 == 0 && (year%100 != 0 || year%400 == 0) { // leap year return 29 } return 28 default: return 31 } } 

EDIT: since I really like measuring things:

 $ go test -bench . testing: warning: no tests to run PASS BenchmarkDim2-8 200000000 7.26 ns/op BenchmarkDim-8 1000000000 2.80 ns/op // LIES! BenchmarkTime-8 10000000 169 ns/op BenchmarkTime2-8 10000000 234 ns/op ok github.com/drathier/scratchpad/go 9.741s 

BenchMarkDim2: not tested, but very fast.

 func daysInMonthTime(month, year int) time.Time { return time.Time{}.Add(time.Hour * 10 + time.Hour*24*30*time.Duration(month-1) + time.Second * time.Duration(daysInMonth(month, year)) * 24 * 60 + 1337) } 

BenchmarkDim: // LIES

 func daysInMonth(month, year int) int { switch time.Month(month) { case time.April, time.June, time.September, time.November: return 30 case time.February: if year%4 == 0 && (year%100 != 0 || year%400 == 0) { // leap year return 29 } return 28 default: return 31 } } 

Benchmarktime:

 func timeDaysInMonth() time.Time { // January, 29th t, _ := time.Parse("2006-01-02", "2016-01-29") y, m, _ := t.Date() lastday := time.Date(y, m+1, 0, 0, 0, 0, 0, time.UTC) return lastday } 

Benchmarktime2

 func time2daysinmonth() time.Time { t, _ := time.Parse("2006-01-02", "2016-01-01") t = t.AddDate(0, 1, 0).AddDate(0, 0, -1) return t } 

Answer 3

In my own code, I used something similar:

 func lastDayOfTheMonth(year, month int) time.Time { if month++; month > 12 { month = 1 } t := time.Date(year, time.Month(month), 0, 0, 0, 0, 0, time.UTC) return t } 

playground

Answer 4

It does not fit, but usually I do the following in any language:

 package main import "fmt" import "time" func main() { fmt.Println("Hello, playground") t, _ := time.Parse("2006-01-02", "2016-01-01") t = t.AddDate(0, 1, 0).AddDate(0,0,-1) fmt.Printf("Last day: %v\n", t) } 

http://play.golang.org/p/JhpOZvEhBw