Parsing an array of json objects in GO

Question

I am trying to parse given API response into a struct. It seems to be an array.

[
   {
      "host_name" : "hostname",
      "perf_data" : "",
      "plugin_output" : "All good",
      "state" : 0
   }
]

I cannot figure out how to create struct for it, I came up with:

type ServiceData struct {
    HostName     string `json:"host_name"`
    PerfData     string `json:"perf_data"`
    PluginOutput string `json:"plugin_output"`
    State        int    `json:"state"`
}

defer resp.Body.Close()
buf := new(bytes.Buffer)
buf.ReadFrom(resp.Body)
jsonStr := buf.String()
var servicedata ServiceData
json.Unmarshal([]byte(jsonStr), &servicedata)

But no luck.

Should I perhaps remove square brackets from the initial response? Could somebody point me in the right direction?

Just want to note that the JSON response is an array of objects containing a single object. You were trying to parse it into a single object and was failing. Creating a variable that is and array of ServiceData fixes the issue as per the answer below. — RisingSun
– RisingSun, Commented Jun 17, 2020 at 22:55

icza · Accepted Answer · 2020-06-18 00:14:35Z

You may unmarshal JSON arrays into Go slices. So unmarshal into a value of type []ServiceData or []*ServiceData:

var servicedata []*ServiceData

A working demo:

func main() {
    var result []*ServiceData
    if err := json.Unmarshal([]byte(src), &result); err != nil {
        panic(err)
    }
    fmt.Printf("%+v", result[0])
}

const src = `[
   {
      "host_name" : "hostname",
      "perf_data" : "",
      "plugin_output" : "All good",
      "state" : 0
   }
]`

Which outputs (try it on the Go Playground):

&{HostName:hostname PerfData: PluginOutput:All good State:0}

Also note that you may unmarshal "directly from the body", no need to read the body first.

Use json.Decoder to do that:

var servicedata []*ServiceData
if err := json.NewDecoder(resp.Body).Decode(&result); err != nil {
    // handle error
}

It's much shorter, easier to read and it's more efficient.

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