自定义将扁平的 JSON 转换为嵌套结构的反序列化方法

huangapple
go评论122阅读模式
英文:

Custom unmarshalling from flat json to nested struct

问题

假设我有两个相关的结构体，如下所示：

  1. type SecretUser struct {
  2.     UserInfo `json:"userInfo"`
  3.     Password string `json:"password"`
  4. }
  6. type UserInfo struct {
  7.     FirstName string `json:"firstName"`
  8.     LastName  string `json:"lastName"`
  9.     Email     string `json:"email"`
  10. }

我收到的JSON数据如下所示：

  1. {
  2.     "firstName": "nice",
  3.     "lastName": "guy",
  4.     "email": "nice@guy.co.uk",
  5.     "password": "abc123"
  6. }

我想将这个JSON数据解组成一个SecretUser结构体。有没有比下面这种方法更好的方式？

  1. func (u *User) UnmarshalJSON(data []byte) error {
  2.     var objmap map[string]*json.RawMessage
  3.     var password string
  4.     var err error
  6.     err = json.Unmarshal(data, &objmap)
  7.     if err != nil {
  8.         return err
  9.     }
  11.     if err := json.Unmarshal(data, &u.UserInfo); err != nil {
  12.         return err
  13.     }
  14.     
  15.     err = json.Unmarshal(*objmap["password"], &password)
  16.     if err != nil {
  17.         return err
  18.     }
  20.     u.Password = password
  21.     return nil
  22. }

基本上，我将JSON数据部分解组成一个UserInfo结构体，然后再次读取它以提取密码。我不想为了清晰地解组这个JSON数据而创建另一个结构体，也不想使用外部库（除非它是标准库的一部分）。有没有更清晰/高效的方法来做到这一点，而不需要两次读取JSON数据或手动从映射中设置每个字段？

英文:

Lets say I have two struct that are related like this:

  1. type SecretUser struct {
  2. 	UserInfo `json:"userInfo"`
  3. 	Password string `json:"password"`
  4. }
  6. type UserInfo struct {
  7. 	FirstName string `json:"firstName"`
  8. 	LastName  string `json:"lastName"`
  9. 	Email     string `json:"email"`
  10. }

And I receive a JSON in this form:

  1. {
  2. 	"firstName": "nice",
  3. 	"lastName":"guy",
  4. 	"email":"nice@guy.co.uk",
  5. 	"password":"abc123"
  6. }

I want to unmarshall this JSON into a SecretUser. Is there a better way than doing it like this?

  1. func (u *User) UnmarshalJSON(data []byte) error {
  2. 	var objmap map[string]*json.RawMessage
  3. 	var password string
  4. 	var err error
  6. 	err = json.Unmarshal(data, &objmap)
  7. 	if err != nil {
  8. 		return err
  9. 	}
  11. 	if err := json.Unmarshal(data, &u.UserInfo); err != nil {
  12. 		return err
  13. 	}
  14. 	
  15. 	err = json.Unmarshal(*objmap["password"], &password)
  16. 	if err != nil {
  17. 		return err
  18. 	}
  20. 	u.Password = password
  21. 	return nil
  22. }

Basically, I partially unmarshall the JSON into a UserInfo struct and then read it again to extract the password. I don't want to create another struct for just unmarshalling this JSON cleanly or use an external library (unless it's part of the standard). Is there a more clean/efficient way of doing this, without reading the JSON twice or setting every field manually from a map?

答案1

得分: 1

只需将UserData包含在SecretUser结构体中，并且不为其指定json标签。

  1. type UserInfo struct {
  2.     FirstName string `json:"firstName"`
  3.     LastName  string `json:"lastName"`
  4.     Email     string `json:"email"`
  5. }
  7. type SecretUser struct {
  8.     UserInfo
  9.     Password string `json:"password"`
  10. }
  12. func main() {
  13.     data := []byte(`{"firstName": "nice","lastName":"guy","email":"nice@guy.co.uk","password":"abc123"}`)
  14.     var u SecretUser
  15.     json.Unmarshal(data, &u)
  16.     fmt.Println(u)
  17. }

Go Play Space 示例

英文:

Simply include the UserData into the SecretUser struct and do not specify a json tag for it.

  1. type UserInfo struct {
  2. 	FirstName string `json:"firstName"`
  3. 	LastName  string `json:"lastName"`
  4. 	Email     string `json:"email"`
  5. }
  7. type SecretUser struct {
  8. 	UserInfo
  9. 	Password string `json:"password"`
  10. }
  12. func main() {
  13. 	data := []byte(`{"firstName": "nice","lastName":"guy","email":"nice@guy.co.uk","password":"abc123"}`)
  14. 	var u SecretUser
  15. 	json.Unmarshal(data, &u)
  16. 	fmt.Println(u)
  17. }

Go Play Space example

huangapple
  • 本文由 发表于 2022年3月4日 19:07:13
  • 转载请务必保留本文链接：https://go.coder-hub.com/71350549.html
  • go
  • json
  • unmarshalling
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