unmarshal generic json in Go [duplicate] - json

Question

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Closed 10 months ago.
I'm a new Go programmer (From Java) and I would like to reproduce a generic way which is esay to use in Java.
I want to create some function which allow me to do an Unmarshal on a JSON string in order to avoid code duplicity.
This is my current code which is not working :
type myStruct1 struct {
id string
name string
}
func (obj myStruct1) toString() string {
var result bytes.Buffer
result.WriteString("id : ")
result.WriteString(obj.id)
result.WriteString("\n")
result.WriteString("name : ")
result.WriteString(obj.name)
return result.String()
}
func main() {
content := `{id:"id1",name="myName"}`
object := myStruct1{}
parseJSON(content, object)
fmt.Println(object.toString())
}
func parseJSON(content string, object interface{}) {
var parsed interface{}
json.Unmarshal([]byte(content), &parsed)
}
This code, on run, returns me this :
id :
name :
Do you have any idea ?
Thanks

Answer 1

The issue is you want to write to a generic type? You probably want a string map. This works with BSON anyways:
var anyJson map[string]interface{}
json.Unmarshal(bytes, &anyJson)
You'll be able to access the fields like so:
anyJson["id"].(string)
Don't forget to type assert your values, and they must be the correct type or they'll panic. (You can read more about type assertions on the golang site)

Answer 2

To parse "generic JSON" when you have no idea what schema it has:
var parsed any
err := json.Unmarshal(jsonText, &parsed)
The returned any in parsed will be a map[string]any or []any or nil or single values float64, bool, string.
You can test the type and react accordingly.
import (
"encoding/json"
"fmt"
)
func test(jsonText []byte) {
// parsing
var parsed any
err := json.Unmarshal(jsonText, &parsed)
if err != nil {
panic(err) // malformed input
}
// type-specific logic
switch val := parsed.(type) {
case nil:
fmt.Println("json specifies null")
case map[string]any:
fmt.Printf("id:%s name:%s\n", val["id"], val["name"])
case []any:
fmt.Printf("list of %d items\n", len(val))
case float64:
fmt.Printf("single number %f\n", val)
case bool:
fmt.Printf("single bool %v\n", val)
case string:
fmt.Printf("single string %s\n", val)
default:
panic(fmt.Errorf("type %T unexpected", parsed))
}
}

Answer 3

Unmarshal will only set exported fields of the struct.
Which means you need to modify the json struct to use capital case letters:
type myStruct1 struct {
Id string
Name string
}
The reason behind this is that the json library does not have the ability to view fields using reflect unless they are exported.

Answer 4

You have to export your fields:
type myStruct1 struct {
Id string
Name string
}
See Exported Identifiers from documentation.

Answer 5

There are a few changes you need to make in your code to make it work:
The function json.Unmarshal can only set variables inside your struct which are exported, that is, which start with capital letters. Use something like ID and Name for your variable names inside myStruct1.
Your content is invalid JSON. What you actually want is {"ID":"id1","Name":"myName"}.
You're passing object to parseJSON but you're using parsed instead, not object. Make parseJSON receive a *myStruct (instead of an interface{}), and use that variable instead of parsed when unmarshalling the string. Also, always handle the error returns, like err := json.Unmarshal(content, object), and check err.
I'd suggest you to do the Golang Tour ;)

Answer 6

You can also set the file as an Object with dynamic properties inside another struct. This will let you add metadata and you read it the same way.
type MyFile struct {
Version string
Data map[string]interface{}
}