Value object to json in go
I have a struct which contains an other struct as value object.
type User struct {
Name string `json:"name"``
Email valueobject.Email `json:"email"`
}
The valueobject.Email looks like this:
type Email struct {
value string
}
func (e *Email) Value() string {
return e.Value
}
I want the value object as an immutable object, there is also a "factory" method in this is not necessary for my problem.
Now I want to return the User struct as json and therefor I use
response := map[string]interface{}{"user": User}
json.NewEncoder(w).Encode(response)
The result is:
{
"user": {
"name": "John Doe",
"email": {
"Email: "johndoe@example.com"
}
}
}
But I want something link this:
{
"user": {
"name": "John Doe",
"email": "johndoe@example.com"
}
}
json go
asked Nov 14 '18 at 7:06
Marvin CasparMarvin Caspar
359317
add a comment |
I have a struct which contains an other struct as value object.
type User struct {
Name string `json:"name"``
Email valueobject.Email `json:"email"`
}
The valueobject.Email looks like this:
type Email struct {
value string
}
func (e *Email) Value() string {
return e.Value
}
I want the value object as an immutable object, there is also a "factory" method in this is not necessary for my problem.
Now I want to return the User struct as json and therefor I use
response := map[string]interface{}{"user": User}
json.NewEncoder(w).Encode(response)
The result is:
{
"user": {
"name": "John Doe",
"email": {
"Email: "johndoe@example.com"
}
}
}
But I want something link this:
{
"user": {
"name": "John Doe",
"email": "johndoe@example.com"
}
}
json go
asked Nov 14 '18 at 7:06
Marvin CasparMarvin Caspar
359317
add a comment |
I have a struct which contains an other struct as value object.
type User struct {
Name string `json:"name"``
Email valueobject.Email `json:"email"`
}
The valueobject.Email looks like this:
type Email struct {
value string
}
func (e *Email) Value() string {
return e.Value
}
I want the value object as an immutable object, there is also a "factory" method in this is not necessary for my problem.
Now I want to return the User struct as json and therefor I use
response := map[string]interface{}{"user": User}
json.NewEncoder(w).Encode(response)
The result is:
{
"user": {
"name": "John Doe",
"email": {
"Email: "johndoe@example.com"
}
}
}
But I want something link this:
{
"user": {
"name": "John Doe",
"email": "johndoe@example.com"
}
}
json go
asked Nov 14 '18 at 7:06
Marvin CasparMarvin Caspar
359317
I have a struct which contains an other struct as value object.
type User struct {
Name string `json:"name"``
Email valueobject.Email `json:"email"`
}
The valueobject.Email looks like this:
type Email struct {
value string
}
func (e *Email) Value() string {
return e.Value
}
I want the value object as an immutable object, there is also a "factory" method in this is not necessary for my problem.
Now I want to return the User struct as json and therefor I use
response := map[string]interface{}{"user": User}
json.NewEncoder(w).Encode(response)
The result is:
{
"user": {
"name": "John Doe",
"email": {
"Email: "johndoe@example.com"
}
}
}
But I want something link this:
{
"user": {
"name": "John Doe",
"email": "johndoe@example.com"
}
}
json go
json go
asked Nov 14 '18 at 7:06
Marvin CasparMarvin Caspar
359317
asked Nov 14 '18 at 7:06
Marvin CasparMarvin Caspar
359317
asked Nov 14 '18 at 7:06
Marvin CasparMarvin Caspar
359317
asked Nov 14 '18 at 7:06
Marvin CasparMarvin Caspar
359317
asked Nov 14 '18 at 7:06
Marvin CasparMarvin Caspar
359317
359317
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
It sounds like you need valueobject.Email to implement the json.Marshaler interface:
func (e *Email) MarshalJSON() (byte, error) {
return json.Marshal(e.Value())
}
That is the bare-minimum to implement what you are asking for. By implementing the json.Marshaler interface, it allows you to customize how json.Marshal renders your value.
Another option is to simplify Email into just a wrapper for string, instead of a struct:
type Email string
func (e Email) Value() string {
return e
}
Since strings are already handled by json.Marshal, it should just work.
answered Nov 14 '18 at 7:25
Dominic BarnesDominic Barnes
23.8k75582
Thanks, the MarshalJSON methods solves my problem
– Marvin Caspar
Nov 14 '18 at 7:47
add a comment |
To make your Email type marshal the way you'd like, you'll need to make it implement the json.marshaler interface. GopherAcademy uses the following example:
func (d Dog) MarshalJSON() (byte, error) {
return json.Marshal(NewJSONDog(d)) }
answered Nov 14 '18 at 7:32
Snake14Snake14
28917
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
It sounds like you need valueobject.Email to implement the json.Marshaler interface:
func (e *Email) MarshalJSON() (byte, error) {
return json.Marshal(e.Value())
}
That is the bare-minimum to implement what you are asking for. By implementing the json.Marshaler interface, it allows you to customize how json.Marshal renders your value.
Another option is to simplify Email into just a wrapper for string, instead of a struct:
type Email string
func (e Email) Value() string {
return e
}
Since strings are already handled by json.Marshal, it should just work.
answered Nov 14 '18 at 7:25
Dominic BarnesDominic Barnes
23.8k75582
Thanks, the MarshalJSON methods solves my problem
– Marvin Caspar
Nov 14 '18 at 7:47
add a comment |
It sounds like you need valueobject.Email to implement the json.Marshaler interface:
func (e *Email) MarshalJSON() (byte, error) {
return json.Marshal(e.Value())
}
That is the bare-minimum to implement what you are asking for. By implementing the json.Marshaler interface, it allows you to customize how json.Marshal renders your value.
Another option is to simplify Email into just a wrapper for string, instead of a struct:
type Email string
func (e Email) Value() string {
return e
}
Since strings are already handled by json.Marshal, it should just work.
answered Nov 14 '18 at 7:25
Dominic BarnesDominic Barnes
23.8k75582
Thanks, the MarshalJSON methods solves my problem
– Marvin Caspar
Nov 14 '18 at 7:47
add a comment |
It sounds like you need valueobject.Email to implement the json.Marshaler interface:
func (e *Email) MarshalJSON() (byte, error) {
return json.Marshal(e.Value())
}
That is the bare-minimum to implement what you are asking for. By implementing the json.Marshaler interface, it allows you to customize how json.Marshal renders your value.
Another option is to simplify Email into just a wrapper for string, instead of a struct:
type Email string
func (e Email) Value() string {
return e
}
Since strings are already handled by json.Marshal, it should just work.
answered Nov 14 '18 at 7:25
Dominic BarnesDominic Barnes
23.8k75582
It sounds like you need valueobject.Email to implement the json.Marshaler interface:
func (e *Email) MarshalJSON() (byte, error) {
return json.Marshal(e.Value())
}
That is the bare-minimum to implement what you are asking for. By implementing the json.Marshaler interface, it allows you to customize how json.Marshal renders your value.
Another option is to simplify Email into just a wrapper for string, instead of a struct:
type Email string
func (e Email) Value() string {
return e
}
Since strings are already handled by json.Marshal, it should just work.
answered Nov 14 '18 at 7:25
Dominic BarnesDominic Barnes
23.8k75582
answered Nov 14 '18 at 7:25
Dominic BarnesDominic Barnes
23.8k75582
answered Nov 14 '18 at 7:25
Dominic BarnesDominic Barnes
23.8k75582
answered Nov 14 '18 at 7:25
Dominic BarnesDominic Barnes
23.8k75582
23.8k75582
Thanks, the MarshalJSON methods solves my problem
– Marvin Caspar
Nov 14 '18 at 7:47
add a comment |
Thanks, the MarshalJSON methods solves my problem
– Marvin Caspar
Nov 14 '18 at 7:47
Thanks, the MarshalJSON methods solves my problem
– Marvin Caspar
Nov 14 '18 at 7:47
Thanks, the MarshalJSON methods solves my problem
– Marvin Caspar
Nov 14 '18 at 7:47
add a comment |
To make your Email type marshal the way you'd like, you'll need to make it implement the json.marshaler interface. GopherAcademy uses the following example:
func (d Dog) MarshalJSON() (byte, error) {
return json.Marshal(NewJSONDog(d)) }
answered Nov 14 '18 at 7:32
Snake14Snake14
28917
add a comment |
To make your Email type marshal the way you'd like, you'll need to make it implement the json.marshaler interface. GopherAcademy uses the following example:
func (d Dog) MarshalJSON() (byte, error) {
return json.Marshal(NewJSONDog(d)) }
answered Nov 14 '18 at 7:32
Snake14Snake14
28917
add a comment |
To make your Email type marshal the way you'd like, you'll need to make it implement the json.marshaler interface. GopherAcademy uses the following example:
func (d Dog) MarshalJSON() (byte, error) {
return json.Marshal(NewJSONDog(d)) }
answered Nov 14 '18 at 7:32
Snake14Snake14
28917
To make your Email type marshal the way you'd like, you'll need to make it implement the json.marshaler interface. GopherAcademy uses the following example:
func (d Dog) MarshalJSON() (byte, error) {
return json.Marshal(NewJSONDog(d)) }
answered Nov 14 '18 at 7:32
Snake14Snake14
28917
edited Nov 14 '18 at 7:43
answered Nov 14 '18 at 7:32
Snake14Snake14
28917
answered Nov 14 '18 at 7:32
Snake14Snake14
28917
answered Nov 14 '18 at 7:32
Snake14Snake14
28917
28917
add a comment |
add a comment |
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