英文:
Get last day in month of time.Time
问题
当我有一个time.Time时：
// 1月29日
t, _ := time.Parse("2006-01-02", "2016-01-29")
我如何得到一个代表1月31日的time.Time？这个例子很简单，但是当有一个日期在二月时，最后一天可能是28号或29号。
英文:
When I have a time.Time:
// January, 29th
t, _ := time.Parse("2006-01-02", "2016-01-29")
How can I get a time.Time which represents January 31st? This example is trivial, but when there's a date in February, the last day might be 28th or 29th.
答案1
得分: 8
> 时间包
>
> func Date
>
> func Date(year int, month Month, day, hour, min, sec, nsec int, loc *Location) Time
>
> Date 函数返回与给定位置中的时间对应的
>
> yyyy-mm-dd hh:mm:ss + nsec 纳秒
>
> 在该时间适当的时区。
>
> 月份、日期、小时、分钟、秒和纳秒的值可能超出其正常范围，在转换过程中将被规范化。例如，10月32日将转换为11月1日。
例如，规范化日期：
package main
import (
"fmt"
"time"
)
func main() {
// 1月29日
t, _ := time.Parse("2006-01-02", "2016-01-29")
fmt.Println(t.Date())
// 1月31日
y,m,_ := t.Date()
lastday:= time.Date(y,m+1,0,0,0,0,0,time.UTC)
fmt.Println(lastday.Date())
}
输出结果：
2016年1月29日
2016年1月31日
英文:
> Package 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 appropriate zone for that time in the given location.
>
> The month, day, hour, min, sec, and nsec values may be outside their
> usual ranges and will be normalized during the conversion. For
> example, October 32 converts 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
答案2
得分: 2
你可以自己编写一个函数，可能是这样的：
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) { // 闰年
return 29
}
return 28
default:
return 31
}
}
编辑：因为我真的很喜欢测量事物：
$ go test -bench .
testing: warning: no tests to run
PASS
BenchmarkDim2-8 200000000 7.26 ns/op
BenchmarkDim-8 1000000000 2.80 ns/op // 谎言！
BenchmarkTime-8 10000000 169 ns/op
BenchmarkTime2-8 10000000 234 ns/op
ok github.com/drathier/scratchpad/go 9.741s
BenchMarkDim2：未经测试，但非常快。
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：谎言
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) {
// 闰年
return 29
}
return 28
default:
return 31
}
}
BenchmarkTime：
func timeDaysInMonth() time.Time {
// 1月29日
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
}
英文:
You could write a function yourself, maybe 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
}
答案3
得分: 1
我已经在我的代码中使用了类似的内容：
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
}
英文:
I've used something similar to this in my own code:
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
}
答案4
得分: 0
这是一个通用的方法，不仅适用于Go语言，通常我在任何语言中都会这样做：
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
英文:
It's not Go specific but usually I do 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)
}
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