字符串2h的持续时间，而不是2h0m0s

Question

默认time.Duration String方法使用0s表示分钟和0m0s表示小时来格式化持续时间。我是否可以使用会产生5m而不是5m0s和2h而不是2h0m0s ...的函数，还是我必须实现自己的函数？

Answer 1

不在标准库中，但是创建一个库很容易：

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
}

正在测试：

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))
}

输出（在Go Playground)上尝试：

5h4m3s   5h4m3s
5h4m0s   5h4m
5h0m3s   5h0m3s
4m3s     4m3s
5h0m0s   5h
4m0s     4m
3s       3s

Answer 2

您可以像这样对持续时间进行四舍五入：

        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()
        }