dsaff82024 2018-10-29 02:47
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I'm attempting to unmarshal a raw json string. There seems to be an error with encoding but I can't quite figure it out.

package main

import (

type Foo struct {
    Transmission string `json:"transmission"`
    Trim         string `json:"trim"`
    Uuid         string `json:"uuid"`
    Vin          string `json:"vin"`

func main() {

    var foo Foo

    sample := `{
        "transmission": "continuously\x20variable\x20automatic",
        "trim": "SL",
        "uuid" : "6993e4090a0e0ae80c59a76326e360a1",
        "vin": "5N1AZ2MH6JN192059"

    err := json.Unmarshal([]byte(sample), &foo)
    if err != nil {


2009/11/10 23:00:00 invalid character 'x' in string escape code

It works if transmission entry is removed.

Here is a working playground.

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2条回答 默认 最新

  • douliao8402 2018-10-29 02:59

    Your input is not valid JSON. The JSON spec states that

    All code points may be placed within the quotation marks except for the code points that must be escaped: quotation mark (U+0022), reverse solidus (U+005C), and the control characters U+0000 to U+001F.

    Additionally, although there are two-character escape sequences, \x is not one of them, and thus it is being correctly interpreted as an invalid escape sequence by the Go parser. If you want to have a backslash literal in your JSON, it needs to be represented by \\ in the JSON input itself. See a modified version of your example: https://play.golang.org/p/JZdPJGpPR5q

    (note that this is not an issue with your Go string literal since you're already using a raw (``) string literal—the JSON itself needs to have two backslashes.)

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
  • douju4278 2018-10-29 14:56

    You can replace \x with \\x using string.Replace() function. Then, Unmarshal the replaced string. Here is a working example.




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