Time duration to string 2h instead 2h0m0s

问题

The default time.Duration String method formats the duration with adding 0s for minutes and 0m0s for hours. Is there function that I can use that will produce 5m instead 5m0s and 2h instead 2h0m0s ... or I have to implement my own?

回答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

回答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()
        }

来源：https://stackoverflow.com/questions/41335155/time-duration-to-string-2h-instead-2h0m0s