Question

我遇到了将“星期几”字符串转换为time.Weekday值的问题。我找不到time包中内置的任何内容。

然后，我编写了这个简单的函数（满足了我的需求）：

var daysOfWeek = [...]string{
    "Sunday",
    "Monday",
    "Tuesday",
    "Wednesday",
    "Thursday",
    "Friday",
    "Saturday",
}

func parseWeekday(v string) (time.Weekday, error) {
    for i := range daysOfWeek {
        if daysOfWeek[i] == v {
            return time.Weekday(i), nil
        }
    }

    return time.Sunday, fmt.Errorf("invalid weekday '%s'", v)
}

还有其他推荐或更多惯用方式在Golang中执行此转换吗？

谢谢！

Answer 1

是的，使用映射而不是数组，因此查找更快，更直接：

var daysOfWeek = map[string]time.Weekday{
    "Sunday":    time.Sunday,
    "Monday":    time.Monday,
    "Tuesday":   time.Tuesday,
    "Wednesday": time.Wednesday,
    "Thursday":  time.Thursday,
    "Friday":    time.Friday,
    "Saturday":  time.Saturday,
}

func parseWeekday(v string) (time.Weekday, error) {
    if d, ok := daysOfWeek[v]; ok {
        return d, nil
    }

    return time.Sunday, fmt.Errorf("invalid weekday '%s'", v)
}

测试：

fmt.Println(parseWeekday("Monday"))
fmt.Println(parseWeekday("Friday"))
fmt.Println(parseWeekday("invalid"))

输出（在Go Playgorund上尝试）：

Monday <nil>
Friday <nil>
Sunday invalid weekday 'invalid'

提示：

您甚至可以使用for循环来安全地初始化daysOfWeek映射，如下所示：

var daysOfWeek = map[string]time.Weekday{}

func init() {
    for d := time.Sunday; d <= time.Saturday; d++ {
        daysOfWeek[d.String()] = d
    }
}

测试和输出是相同的。在Go Playground上尝试这个。

此映射解决方案的另一个不错的属性（与数组解决方案相比）是，您可以在同一映射中列出可以解析为time.Weekday的其他有效值，而无需其他解析代码。

例如，让我们还将3个字母的简短工作日名称解析为与它们相等的time.Weekday，例如"Mon"至time.Monday。

可以通过一个简单的循环添加此扩展名：

var daysOfWeek = map[string]time.Weekday{}

func init() {
    for d := time.Sunday; d <= time.Saturday; d++ {
        name := d.String()
        daysOfWeek[name] = d
        daysOfWeek[name[:3]] = d
    }
}

测试：

fmt.Println(parseWeekday("Monday"))
fmt.Println(parseWeekday("Friday"))
fmt.Println(parseWeekday("Mon"))
fmt.Println(parseWeekday("Fri"))
fmt.Println(parseWeekday("invalid"))

输出（在Go Playground上尝试）：

Monday <nil>
Friday <nil>
Monday <nil>
Friday <nil>
Sunday invalid weekday 'invalid'

Answer 2

这看起来可以做到：

package main
import "time"

func parseWeekday(v string) (time.Weekday, error) {
   t, e := time.Parse("Monday 2", v + " 2")
   if e != nil { return 0, e }
   return t.Weekday(), nil
}

func main() {
   n, e := parseWeekday("Sunday")
   if e != nil {
      panic(e)
   }
   println(n == time.Sunday)
}

https://golang.org/pkg/time#Parse

将Weekday字符串解析为时间。

2 个答案: