json.Unmarshal 似乎不关注结构标签

Question

我有一个如下所示的 JSON 对象:

{"API version":"1.2.3"}

我想使用 json.Unmarshal() 将其转换为对象go 中的函数。根据this blog post :

How does Unmarshal identify the fields in which to store the decoded data? For a given JSON key "Foo", Unmarshal will look through the destination struct's fields to find (in order of preference):

• An exported field with a tag of "Foo" (see the Go spec for more on struct tags),
• An exported field named "Foo", or
• An exported field named "FOO" or "FoO" or some other case-insensitive match of "Foo".

unmarshal documentation 证实了这一点.

由于“API版本”中有一个空格，这不是一个有效的go标识符，所以我在该字段上使用了一个标签:

type ApiVersion struct {
 Api_version string "API version"
}

我尝试像这样解码它:

func GetVersion() (ver ApiVersion, err error) {

 // Snip code that gets the JSON

 log.Println("Json:",string(data))
 err = json.Unmarshal(data,&ver)
 log.Println("Unmarshalled:",ver);
}

输出为:

2014/01/06 16:47:38 Json: {"API version":"1.2.3"}
2014/01/06 16:47:38 Unmarshalled: {}

如您所见，JSON 没有被编码到 ver 中。我错过了什么？

Answer 1

encoding/json 模块要求结构体标签具有命名空间。所以你想要类似的东西:

type ApiVersion struct {
    Api_version string `json:"API version"`
}

这样做是为了使 json 结构标记可以与其他库(例如 XML 编码器)中的标记共存。