Using EasyJSON with golang

11,846

Solution 1

I don't know why you trying to use easyjson. encoding/json is pretty fine to work with. But though here is the answer for you.

NB: It would be better if you use encoding/json.

//easyjson:json
type JSONData struct {
    Data []string
}

After define this struct run easyjson <fileName-JSONData-is-defined>.go. this will create an extra go file containg

func (v JSONData) MarshalJSON() ([]byte, error)
func (v JSONData) MarshalEasyJSON(w *jwriter.Writer)
func (v *JSONData) UnmarshalJSON(data []byte) errorfunc (v *JSONData) 
func UnmarshalEasyJSON(l *jlexer.Lexer)

those methods. Then (un-)marshal using

d := &JSONData{}
d.UnmarshalJSON([]byte(`{"Data" : ["One", "Two", "Three"]} `))
// Or you could also use
// json.Unmarshal(data, d) this will also call this d.UnmarshalJSON
fmt.Println(d)

A full example is here.

Solution 2

Well, easyJson is 4 times faster than normal json(as per its documets) ,in our organization we have used it extensively and yes its faster. Here is a small example to get started. my current directory name is easyJson

vim easyjson.go

package main

import "fmt"
import "time"
import ej "random/golang/easyJson/model"

func main() {
 t1 := time.Now()
 var d ej.Data
 d.Name = "sharathbj"
 d.Age = 23
 data, _ := d.MarshalJSON()
 fmt.Println(string(data))
 fmt.Println("elapsedTime:", time.Now().Sub(t1))
}

create a directory named model ,where your structures are defined and new go file models.go

mkdir model

vim models.go

package easyJson

//easyjson:json
type Data struct {
  Name string `json:"name"`
  Age  int    `json:"age"`
}

Now run command to create a easyjson file (-all specified to reference all the structure inside a given file)

easyjson -all model/models.go

now a new file will be generated models_easyjson.go using which marshaling/unmarshaling will be referenced

go run easyjson.go

enter image description here

To compare easyjson with normal encoding/json ,below is the code

vim normaljson.go

package main

import (
   "fmt"
   "time"
   "encoding/json"
    model "random/golang/easyJson/model"
 )
 func main() {
   t1 := time.Now()
   var d model.Data
   d.Name = "sharathbj"
   d.Age = 23
   data, _ := json.Marshal(d)
   fmt.Println(string(data))
   fmt.Println("elapsedTime:", time.Now().Sub(t1))
 }

enter image description here

Clearly easyjson is 7 micro second faster than normal json ,and you will see its impact when you do it for bigger structures , u can see the source code below .
https://github.com/sharathbj/random/tree/master/golang/easyJson

Cheers!!

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11,846
tuk
Author by

tuk

Updated on June 04, 2022

Comments

  • tuk
    tuk almost 2 years

    Let's say I have a struct like below:-

    //easyjson:json
    type JSONData struct {
        Data []string
    }
    

    I want to un-marshal the below json to JSONData struct

    {"Data" : ["One", "Two", "Three"]} 
    

    Can someone let me know how can I use easyjson to un-marshal a json in Golang? I could not find any example in their README