如何在Golang中响应所有包含标记为temp的字段（包括字段）？

In my Webservice, I have a model:

// Comment struct
type Comment struct {
    Owner            UserObject      `json:"owner"`
    ID               int64           `json:"id"`
    Message          string          `json:"message"`
    Mentions         []MentionObject `json:"mentions,omitempty"`
    CreatedAt        int64           `json:"created_at,omitempty"`
    UpdatedAt        int64           `json:"updated_at,omitempty"`
    Status           int             `json:"status,omitempty"`
    CanEdit          bool            `json:"can_edit"`
    CanDelete        bool            `json:"can_delete"`
}


// UserObject struct
type UserObject struct {
    ID       int64  `json:"id"`
    Username string `json:"username"`
    FullName string `json:"full_name"`
    Avatar   string `json:"avatar"`
}

// MentionObject struct
type MentionObject struct {
    ID     int64 `json:"id"`
    Length int64 `json:"length"`
    Offset int64 `json:"offset"`
}

I have use gin gonic to routing

routes.GET("/user", func(c *gin.Context) {
        c.JSON(200, Comment{})
    })

I need return all fields of this struct, I know that It going to response a json:

{
  "owner": {
    "id": 0,
    "username": "",
    "full_name": "",
    "avatar": ""
  },
  "id": 0,
  "message": "",
  "can_report": false,
  "can_edit": false,
  "can_delete": false
}

I know that This is right, but I still want response all of field. How to do that?

If you don't want to drop the omitempty value from the tag because you need it for some other purpose, you can either, if you're on Go 1.8+, define a new type identical to the one you want to serialize but without the omitempty tag values and then simply convert a value from the old type to the new type.

Here's an example of the 1.8+ "tag ignoring" type conversion

You can also define a new type with only those fields that are omitted in the original and then embed the original in the new type like so.