How to decode JSON with type convert from string to float64

Question

I need to decode a JSON string with the float number like:

{"name":"Galaxy Nexus", "price":"3460.00"}

I use the Golang code below:

package main

import (
    "encoding/json"
    "fmt"
)

type Product struct {
    Name  string
    Price float64
}

func main() {
    s := `{"name":"Galaxy Nexus", "price":"3460.00"}`
    var pro Product
    err := json.Unmarshal([]byte(s), &pro)
    if err == nil {
        fmt.Printf("%+v\n", pro)
    } else {
        fmt.Println(err)
        fmt.Printf("%+v\n", pro)
    }
}

When I run it, get the result:

json: cannot unmarshal string into Go value of type float64
{Name:Galaxy Nexus Price:0}

I want to know how to decode the JSON string with type convert.

Dustin · Accepted Answer · 2012-03-05 21:01:53Z

236

The answer is considerably less complicated. Just add tell the JSON interpeter it's a string encoded float64 with ,string (note that I only changed the Price definition):

package main

import (
    "encoding/json"
    "fmt"
)

type Product struct {
    Name  string
    Price float64 `json:",string"`
}

func main() {
    s := `{"name":"Galaxy Nexus", "price":"3460.00"}`
    var pro Product
    err := json.Unmarshal([]byte(s), &pro)
    if err == nil {
        fmt.Printf("%+v\n", pro)
    } else {
        fmt.Println(err)
        fmt.Printf("%+v\n", pro)
    }
}

answered Mar 5, 2012 at 21:01

Dustin

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9 Comments

yanunon Over a year ago

Thank you! I think this is the best solution for my problem. Could you tell me where is the official document about the usage of ",string"?

Darrrrrren Over a year ago

Note that the leading comma in front of json:",string" is necessary - it won't work without it.

Kamil Dziedzic Over a year ago

@dustin any idea how to do that other way around? Convert json number to go string? e.g. "N": 1234 to N: "1234"?

dao leno Over a year ago

Official document: golang.org/pkg/encoding/json/#Marshal

The "string" option signals that a field is stored as JSON inside a JSON-encoded string. It applies only to fields of string, floating point, integer, or boolean types. This extra level of encoding is sometimes used when communicating with JavaScript programs

aolivera Over a year ago

Does anybody know how to accept both number and string?

|

Salvador Dali · Accepted Answer · 2016-08-02 20:26:46Z

22

Just letting you know that you can do this without Unmarshal and use json.decode. Here is Go Playground

package main

import (
    "encoding/json"
    "fmt"
    "strings"
)

type Product struct {
    Name  string `json:"name"`
    Price float64 `json:"price,string"`
}

func main() {
    s := `{"name":"Galaxy Nexus","price":"3460.00"}`
    var pro Product
    err := json.NewDecoder(strings.NewReader(s)).Decode(&pro)
    if err != nil {
        fmt.Println(err)
        return
    }
    fmt.Println(pro)
}

edited Aug 2, 2016 at 20:26

user6169399

answered Nov 19, 2015 at 7:38

Salvador Dali

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2 Comments

theNoble247 Over a year ago

Is there a way to do the reverser? i.e decode a number into a string variable?

Yuseferi Over a year ago

@theNoble247 did you find a solution for it? ( reverser)

g10guang · Accepted Answer · 2018-10-23 07:11:10Z

6

Avoid converting a string to []byte: b := []byte(s). It allocates a new memory space and copy the whole the content into it.

strings.NewReader interface is better. Below is the code from godoc:

package main

import (
    "encoding/json"
    "fmt"
    "io"
    "log"
    "strings"
)

func main() {
    const jsonStream = `
    {"Name": "Ed", "Text": "Knock knock."}
    {"Name": "Sam", "Text": "Who's there?"}
    {"Name": "Ed", "Text": "Go fmt."}
    {"Name": "Sam", "Text": "Go fmt who?"}
    {"Name": "Ed", "Text": "Go fmt yourself!"}
`
    type Message struct {
        Name, Text string
    }
    dec := json.NewDecoder(strings.NewReader(jsonStream))
    for {
        var m Message
        if err := dec.Decode(&m); err == io.EOF {
            break
        } else if err != nil {
            log.Fatal(err)
        }
        fmt.Printf("%s: %s\n", m.Name, m.Text)
    }
}

answered Oct 23, 2018 at 7:11

g10guang

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1 Comment

user12817546 Over a year ago

Your answer makes a great point that I quoted. Did I add to it? See stackoverflow.com/a/62740786/12817546

Mostafa · Accepted Answer · 2012-02-26 12:45:55Z

4

Passing a value in quotation marks make that look like string. Change "price":"3460.00" to "price":3460.00 and everything works fine.

If you can't drop the quotations marks you have to parse it by yourself, using strconv.ParseFloat:

package main

import (
    "encoding/json"
    "fmt"
    "strconv"
)

type Product struct {
    Name       string
    Price      string
    PriceFloat float64
}

func main() {
    s := `{"name":"Galaxy Nexus", "price":"3460.00"}`
    var pro Product
    err := json.Unmarshal([]byte(s), &pro)
    if err == nil {
        pro.PriceFloat, err = strconv.ParseFloat(pro.Price, 64)
        if err != nil { fmt.Println(err) }
        fmt.Printf("%+v\n", pro)
    } else {
        fmt.Println(err)
        fmt.Printf("%+v\n", pro)
    }
}

answered Feb 26, 2012 at 12:45

Mostafa

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3 Comments

yanunon Over a year ago

Is there a way that not changing the 'Product' struct and implement a decode function or interface to parse the string to float?

Mostafa Over a year ago

@yanunon Yes, you can use a map of type map[string]interface{} for Unmarshal, and parse that into your struct.

Mostafa Over a year ago

@yanunon Or if you want real flexibility, you can write your own Unmarshal, which calls default Unmarshal with a map[string]interface{}, but uses reflect and strconv packages to do the parsing.

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