如何在Golang中响应所有包含标记为temp的字段（包括字段）？

Question

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?

Answer 1

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.