Time duration to string 2h instead 2h0m0s
13,592
Solution 1
Foreword: I released this utility in github.com/icza/gox
, see timex.ShortDuration()
.
Not in the standard library, but it's really easy to create one:
func shortDur(d time.Duration) string {
s := d.String()
if strings.HasSuffix(s, "m0s") {
s = s[:len(s)-2]
}
if strings.HasSuffix(s, "h0m") {
s = s[:len(s)-2]
}
return s
}
Testing it:
h, m, s := 5*time.Hour, 4*time.Minute, 3*time.Second
ds := []time.Duration{
h + m + s, h + m, h + s, m + s, h, m, s,
}
for _, d := range ds {
fmt.Printf("%-8v %v\n", d, shortDur(d))
}
Output (try it on the Go Playground):
5h4m3s 5h4m3s
5h4m0s 5h4m
5h0m3s 5h0m3s
4m3s 4m3s
5h0m0s 5h
4m0s 4m
3s 3s
Solution 2
You can just round the duration like this:
func RountTime(roundTo string, value time.Time) string {
since := time.Since(value)
if roundTo == "h" {
since -= since % time.Hour
}
if roundTo == "m" {
since -= since % time.Minute
}
if roundTo == "s" {
since -= since % time.Second
}
return since.String()
}
Author by
gsf
Updated on June 08, 2022Comments
-
gsf almost 2 years
The default time.Duration String method formats the duration with adding
0s
for minutes and0m0s
for hours. Is there function that I can use that will produce5m
instead5m0s
and2h
instead2h0m0s
... or I have to implement my own?