Golang解析时间.持续时间

Question

我想解析一下time.Duration.持续时间是"PT15M"(字符串/字节),并希望将其转换为有效time.Duration.

---

如果这是一time.Time件事,我会用:

t, err := time.Parse(time.RFC3339Nano, "2013-06-05T14:10:43.678Z")

但这不存在(ParseDuration只需要一个参数):

d, err := time.ParseDuration(time.RFC3339Nano, "PT15M")

---

我如何解析ISO 8601持续时间？

Answer 1

它并不完全是“开箱即用”，但正则表达式可以完成这项工作：

package main

import "fmt"
import "regexp"
import "strconv"
import "time"

func main() {
    fmt.Println(ParseDuration("PT15M"))
    fmt.Println(ParseDuration("P12Y4MT15M"))
}

func ParseDuration(str string) time.Duration {
    durationRegex := regexp.MustCompile(`P(?P<years>\d+Y)?(?P<months>\d+M)?(?P<days>\d+D)?T?(?P<hours>\d+H)?(?P<minutes>\d+M)?(?P<seconds>\d+S)?`)
    matches := durationRegex.FindStringSubmatch(str)

    years := ParseInt64(matches[1])
    months := ParseInt64(matches[2])
    days := ParseInt64(matches[3])
    hours := ParseInt64(matches[4])
    minutes := ParseInt64(matches[5])
    seconds := ParseInt64(matches[6])

    hour := int64(time.Hour)
    minute := int64(time.Minute)
    second := int64(time.Second)
    return time.Duration(years*24*365*hour + months*30*24*hour + days*24*hour + hours*hour + minutes*minute + seconds*second)
}

func ParseInt64(value string) int64 {
    if len(value) == 0 {
        return 0
    }
    parsed, err := strconv.Atoi(value[:len(value)-1])
    if err != nil {
        return 0
    }
    return int64(parsed)
}