Unmarshalling JSON string with Golang

Question

I have the following json string

{"x":{"l.a":"test"}}

type Object struct {
    Foo  map[string]map[string]string `json:"l.a"`
    }
     var obj Object
    err = json.Unmarshal(body, &obj)
    if err != nil{
        fmt.Println(err)
    }
    fmt.Println("jsonObj", obj)

but get empty any idea how can i get the string "test"

You're defining Foo to look into the l.a property but that property is not present in the root object. Maybe your intention is to have json:"x" as the tag. Also, map[string]map[string]string is redundant, Object acts like the root object so you would need just map[string]string. — David Lilue
– David Lilue, Commented Sep 9, 2022 at 1:59

Brits · Accepted Answer · 2022-09-09 02:06:52Z

2

With your code Unmarshal is looking for l.a at the top level in the JSON (and it's not there - x is).

There are a number of ways you can fix this, the best is going to depend upon your end goal; here are a couple of them (playground)

const jsonTest = `{"x":{"l.a":"test"}}`

type Object struct {
    Foo map[string]string `json:"x"`
}

func main() {
    var obj Object
    err := json.Unmarshal([]byte(jsonTest), &obj)
    if err != nil {
        fmt.Println(err)
    }
    fmt.Println("jsonObj", obj)

    var full map[string]map[string]string
    err = json.Unmarshal([]byte(jsonTest), &full)
    if err != nil {
        fmt.Println(err)
    }
    fmt.Println("jsonObj2", full)

}

(Got a phone call while entering this and see @DavidLilue has provided a similar comment but may as well post this).

answered Sep 9, 2022 at 2:06

Brits

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0x4e696b68696c · Accepted Answer · 2022-09-09 05:37:10Z

1

You can always create a struct to unmarshall your json

type My_struct struct {
    X struct {
        LA string `json:"l.a"`
    } `json:"x"`
}

func main() {
    my_json := `{"x":{"l.a":"test"}}`
    var obj My_struct
    json.Unmarshal([]byte(my_json), &obj)
    fmt.Println(obj.X.LA)
}

here you are creating a struct and then unmarshalling your json string to its object, so when you do obj.X.LA you will get the string

answered Sep 9, 2022 at 5:37

0x4e696b68696c

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