从字符串获取整数月份值 [英] Get integer month value from string

问题描述

我正在从AWS解析一个cron字符串，它看起来像这样的 cron(0 7 November 13?2019).有没有一种干净的方法，可以使用Go的内置类型从 11月返回到 11 ? time.Month 类型允许将 int 映射到 string ，但是似乎没有相反的方法.我想念什么吗?现在，我已经编写了此代码，以获取正在使用的 map [string] int : monthi:= getMonths()[monthName] .

I'm parsing a cron string from AWS that looks like this cron(0 7 13 November ? 2019). Is there a clean way to go from November back to 11 using Go's built in types? The time.Month type allows mapping int to string, but there doesn't seem to be a way to do the reverse. Am I missing something? For now, I've written this to get a map[string]int that I'm using like this: monthi := getMonths()[monthName].

func getMonths() map[string]int {
    m := make(map[string]int)
    for i := 1; i < 13; i++ {
        month := time.Month(i).String()
        m[month] = i
    }
    return m
}

推荐答案

前言:我在 timex.ParseMonth() .

这是目前最好的方法.

最好是使用包级变量，并仅填充一次地图.

Best would be to use a package level variable and populate the map only once.

通过这种方式进行人口调查更加干净和安全:

And it's much cleaner and safer to do the population this way:

var months = map[string]time.Month{}

func init() {
    for i := time.January; i <= time.December; i++ {
        months[i.String()] = i
    }
}

测试:

for _, s := range []string{"January", "December", "invalid"} {
    m := months[s]
    fmt.Println(int(m), m)
}

输出(在游乐场上尝试):

1 January
12 December
0 %!Month(0)

请注意，此映射具有灵活性，您可以添加短月份名称，以映射到同一月份.例如.您还可以添加 months ["Jan"] = time.January ，因此，如果您输入的是"Jan" ，您还可以获取时间.一月.可以通过在同一循环中将长名称切片来轻松完成此操作，例如:

Note that this map has the flexibility that you may add short month names, mapping to the same month. E.g. you may also add months["Jan"] = time.January, so if your input is "Jan", you would also be able to get time.January. This could easily be done by slicing the long name, in the same loop, for example:

for i := time.January; i <= time.December; i++ {
    name := i.String()
    months[name] = i
    months[name[:3]] = i
}

还请注意，可以使用 time.Parse() 在布局字符串为"January" 的情况下进行解析:

Also note that it's possible to use time.Parse() to do the parsing where the layout string is "January":

for _, s := range []string{"January", "December", "invalid"} {
    t, err := time.Parse("January", s)
    m := t.Month()
    fmt.Println(int(m), m, err)
}

哪些输出(在进入游乐场上尝试):

Which outputs (try it on the Go Playground):

1 January <nil>
12 December <nil>
1 January parsing time "invalid" as "January": cannot parse "invalid" as "January"

但是简单的地图查找在性能上胜于此.

But the simple map lookup is superior to this in performance.

看到类似的问题:将Weekday字符串解析为time.Weekday

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