json - 将 JSON 解码为 UUID 类型

Question

我正在尝试使用 json.Unmarshaler 接口(interface)将 UUID 解码为结构上的 uuid.UUID 字段。我创建了一个名为 myUUID 的自定义类型，一切正常，直到我尝试访问通常位于 uuid.UUID 上的方法。我该如何处理？我是 Go 的新手，所以我可能还没有完全理解自定义类型。

package main

import (
    "encoding/json"
    "errors"
    "fmt"

    "code.google.com/p/go-uuid/uuid"
)

var jsonstring = `
{
    "uuid": "273b62ad-a99d-48be-8d80-ccc55ef688b4"
}
`

type myUUID uuid.UUID

type Data struct {
    uuid myUUID
}

func (u *myUUID) UnmarshalJson(b []byte) error {
    id := uuid.Parse(string(b[:]))
    if id == nil {
            return errors.New("Could not parse UUID")
    }
    *u = myUUID(id)
    return nil
}

func main() {
    d := new(Data)
    err := json.Unmarshal([]byte(jsonstring), &d)
    if err != nil {
            fmt.Printf("%s", err)
    }
    fmt.Println(d.uuid.String())
}

Answer 1

您可能想确保您的 myuuid 变量在 Data struct 中可见/导出:如在“public”中一样。
类型别名 MyUUID 相同(而不是 myUUID)

type MyUUID uuid.UUID

type Data struct {
    Uuid MyUUID
}

来自 JSON and Go :

The json package only accesses the exported fields of struct types (those that begin with an uppercase letter).

---

作为commented通过 Ainar G , style guide还推荐:

Words in names that are initialisms or acronyms (e.g. "URL" or "NATO") have a consistent case.
For example, "URL" should appear as "URL" or "url" (as in "urlPony", or "URLPony"), never as "Url". Here's an example: ServeHTTP not ServeHttp.

This rule also applies to "ID" when it is short for "identifier," so write "appID" instead of "appId".

在您的情况下，这意味着:

type Data struct {
    UUID MyUUID
}