如何在Golang的map中检查三个不同的结构键?

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英文:

How to check for three different struct keys in the golang map?

问题

我有一个方法,在这个方法中,我创建了三个不同的结构体对象,将它们作为键来检查我的name映射。如果我能找到值,我就返回它,否则返回空字符串和状态。

func (r *clientRepo) GetName(id int64, name map[models.LocaleInfo]string, clientId int32, code models.Locale) (string, bool) {
    localeInfo := models.LocaleInfo{
        Locale:  code,
        GroupID: int(clientId),
    }

    localeInfoUS := models.LocaleInfo{
        Locale:  "US",
        GroupID: int(clientId),
    }

    localeInfoGB := models.LocaleInfo{
        Locale:  "GB",
        GroupID: int(clientId),
    }

    value := ""
    ok := false
    if value, ok = name[localeInfo]; !ok {
        if value, ok = name[localeInfoUS]; !ok {
            if value, ok := name[localeInfoGB]; !ok {
                errs.Newf("Unable to find locale for ProductId: %d", id)
                return value, ok
            }
        }
    }
    
    return value, ok
}

有没有办法改进上面的代码,不需要为每个US和GB的区域信息保留单独的行。我的if块看起来也很嵌套,我认为这方面也可以有很大的改进。

英文:

I have a below method where I create three different struct object to check as a key in my name map. If I can find the value, I return back otherwise I return empty string and status.

func (r *clientRepo) GetName(id int64, name map[models.LocaleInfo]string, clientId int32, code models.Locale) (string, bool) {
	localeInfo := models.LocaleInfo{
		Locale:  code,
		GroupID: int(clientId),
	}

	localeInfoUS := models.LocaleInfo{
		Locale:  "US",
		GroupID: int(clientId),
	}

	localeInfoGB := models.LocaleInfo{
		Locale:  "GB",
		GroupID: int(clientId),
	}

	value := ""
	ok := false
	if value, ok = name[localeInfo]; !ok {
		if value, ok = name[localeInfoUS]; !ok {
			if value, ok := name[localeInfoGB]; !ok {
				errs.Newf("Unable to find locale for ProductId: %d", id)
				return value, ok
			}
		}
	}
	
	return value, ok
}

Is there any way to improve above code where I don't have to keep separate line for each US and GB locale info. My if block also looks very nested and I think that can be improved a lot as well.

答案1

得分: 1

特殊的“comma-ok”形式只能在赋值语句中使用,所以无法使其更紧凑。

你可以编写一个实用函数,使用for循环来检查任意数量的键。

使用Go 1.18泛型,你可以使其适用于任何映射类型:

func findKeys[K comparable, V any](m map[K]V, keys ...K) (v V, ok bool) {
    for _, key := range keys {
        if v, ok = m[key]; ok {
            return
        }
    }
    return
}

测试一下:

m := map[int]string{
    1: "one",
    2: "two",
    3: "three",
    4: "four",
}

fmt.Println(findKeys(m, 5, 6, 4))
fmt.Println(findKeys(m, 1, 2, 6))
fmt.Println(findKeys(m, 6, 8))

输出结果为(在Go Playground上尝试一下):

four true
one true
 false

注意:如果你使用的是Go 1.18之前的版本,只需使用非泛型版本的findKeys()

func findKeys(m map[models.LocaleInfo]string, keys ...models.LocaleInfo) (v string, ok bool) {
    for _, key := range keys {
        if v, ok = m[key]; ok {
            return
        }
    }
    return
}

你可以像这样使用findKeys()

value, ok := findKeys(name, localeInfo, localeInfoUS, localeInfoGB)
if !ok {
    errs.Newf("Unable to find locale for ProductId: %d", id)
}
return value, ok

还要注意,如果你使用的是非泛型版本,由于它将被限制为你的映射类型,你可以将更多的代码移到其中:它可以处理键的创建,因此你只需要传递string类型的区域设置值(以及clientId):

func findKeys(names map[models.LocaleInfo]string, clientId int, locales ...string) (v string, ok bool) {
    key := models.LocaleInfo{GroupID: clientId}
    for _, locale := range locales {
        key.Locale = locale
        if v, ok = names[key]; ok {
            return
        }
    }
    return
}

调用它的方式如下:

value, ok := findKeys(names, int(clientId), code, "US", "GB")

参考问题:https://stackoverflow.com/questions/41978101/check-if-key-exists-in-multiple-maps-in-one-condition/41978277#41978277

英文:

The special "comma-ok" form can only be used in an assignment, so you can't make it more compact.

What you may do is write a utility function which uses a for loop to check any number of keys.

Using Go 1.18 generics, you can make it work with any map type:

func findKeys[K comparable, V any](m map[K]V, keys ...K) (v V, ok bool) {
	for _, key := range keys {
		if v, ok = m[key]; ok {
			return
		}
	}
	return
}

Testing it:

m := map[int]string{
	1: "one",
	2: "two",
	3: "three",
	4: "four",
}

fmt.Println(findKeys(m, 5, 6, 4))
fmt.Println(findKeys(m, 1, 2, 6))
fmt.Println(findKeys(m, 6, 8))

Which outputs (try it on the Go Playground):

four true
one true
 false

Note: if you're using a Go prior to Go 1.18, simply use a non-generic version of findKeys():

func findKeys(m map[models.LocaleInfo]string, keys ...models.LocaleInfo) (v string, ok bool) {
    for _, key := range keys {
        if v, ok = m[key]; ok {
            return
        }
    }
    return
}

You could use this findKeys() like this:

value, ok := findKeys(name, localeInfo, localeInfoUS, localeInfoGB)
if !ok {
	errs.Newf("Unable to find locale for ProductId: %d", id)
}
return value, ok

Also note that if you're using a non-generic version, since it will be restricted to your map type anyway, you can move more code into it: it can handle the key creation, so you only have to pass the string locale values (and the clientId):

func findKeys(names map[models.LocaleInfo]string, clientId int, locales ...string) (v string, ok bool) {
	key := models.LocaleInfo{GroupID: clientId}
	for _, locale := range locales {
		key.Locale = locale
		if v, ok = names[key]; ok {
			return
		}
	}
	return
}

Calling it is like:

value, ok := findKeys(names, int(clientId), code, "US", "GB")

See related question: https://stackoverflow.com/questions/41978101/check-if-key-exists-in-multiple-maps-in-one-condition/41978277#41978277

huangapple
  • 本文由 发表于 2022年3月18日 15:13:12
  • 转载请务必保留本文链接:https://go.coder-hub.com/71523520.html
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