从json合并两个map [string] interface {}

Question

I have two json inputs built this way

"count: 1 result: fields"   

I would like to concatenate the fields that I find within result without using a defined structure. I have tried in many ways but most of the time the result is an error about the type Interface {} or the last map overwritten the data

I would like both the "result" and the first and second map fields to be merged within the result in output.

oracle, err := http.Get("http://XXX:8080/XXXX/"+id)
    if err != nil {
            panic(err)
    }
defer oracle.Body.Close()

mysql, err := http.Get("http://XXX:3000/XXX/"+id)
if err != nil {
        panic(err)
}
defer mysql.Body.Close()

oracleJSON, err := ioutil.ReadAll(oracle.Body)
if err != nil {
        panic(err)
}
mysqlJSON, err := ioutil.ReadAll(mysql.Body)
if err != nil {
        panic(err)
}

var oracleOUT map[string]interface{}
var mysqlOUT map[string]interface{}

json.Unmarshal(oracleJSON, &oracleOUT)
json.Unmarshal(mysqlJSON, &mysqlOUT)

a := oracleOUT["result"]
b := mysqlOUT["result"]


c.JSON(http.StatusOK, gin.H{"result": ????})

this is an example of json

{"count":1,"result":{"COD_DIPENDENTE":"00060636","MATRICOLA":"60636","COGNOME":"PIPPO"}}

If i have two json like this the result of the function it should be

`"result":{"COD_DIPENDENTE":"00060636","MATRICOLA":"60636","COGNOME":"PIPPO","COD_DIPENDENTE":"00060636","MATRICOLA":"60636","COGNOME":"PIPPO"}}` 

Answer 1

The output you are looking for is not valid JSON. However with a small change you can output something very similar to your example that is valid JSON.

You probably do want to use a defined structure for the portion of the input that has a known structure, so that you can extract the more abstract "result" section more easily.

If you start at the top of the input structure using a map[string]interface{} then you'll have to do a type assertion on the "result" key. For example:

var input map[string]interface{}

err = json.Unmarshal(data, &input)
if err != nil {
    return err
}
keys, ok := input["result"].(map[string]interface{})
if !ok {
    return errors.New("wasn't the type we expected")
}

However if you used a defined structure for the top level you can do it like the following which feels much cleaner.

type Input struct {
    Count  int                    `json:"count"`
    Result map[string]interface{} `json:"result"`
}

var input Input
err = json.Unmarshal(data, &input)
if err != nil {
    return err
}
// from here you can use input.Result directly without a type assertion

To generate output that has duplicate keys, you could use an array of objects with a single key/value pair in each, then you end up with a valid JSON structure that does not overwrite keys. Here's how to do that (playground link):

package main

import (
    "encoding/json"
    "fmt"
)

type Input struct {
    Count  int                    `json:"count"`
    Result map[string]interface{} `json:"result"`
}

type Output struct {
    Count  int                      `json:"count"`
    Result []map[string]interface{} `json:"result"`
}

var inputdata = [][]byte{
    []byte(`{"count":1,"result":{"COD_DIPENDENTE":"00060636", "MATRICOLA":"60636", "COGNOME":"PIPPO"}}`),
    []byte(`{"count":1,"result":{"COD_DIPENDENTE":"00060636", "MATRICOLA":"60636", "COGNOME":"PIPPO"}}`),
}

func main() {
    inputs := make([]Input, len(inputdata))
    for i := range inputs {
        err := json.Unmarshal(inputdata[i], &inputs[i])
        if err != nil {
            panic(err)
        }
    }

    var out Output
    out.Count = len(inputs)
    for _, input := range inputs {
        for k, v := range input.Result {
            out.Result = append(out.Result, map[string]interface{}{k: v})
        }
    }

    outdata, _ := json.Marshal(out)
    fmt.Println(string(outdata))
}

Which produces output that looks like this when formatted:

{
  "count": 2,
  "result": [
    {"MATRICOLA": "60636"},
    {"COGNOME": "PIPPO"},
    {"COD_DIPENDENTE": "00060636"},
    {"COGNOME": "PIPPO"},
    {"COD_DIPENDENTE": "00060636"},
    {"MATRICOLA": "60636"}
  ]
}