我的结构体无法转换为json。

huangapple
go评论116阅读模式
英文:

My structures are not marshalling into json

问题

我正在使用Go 1.0.3在Mac OS X 10.8.2上，并且正在尝试使用json包将结构体编组为json，但是我一直得到一个空的{} json对象。

根据json.Marshal函数，err值为nil，所以没有任何问题，结构体也是正确的。为什么会发生这种情况？

  1. package main
  3. import (
  4.   "encoding/json"
  5.   "fmt"
  6. )
  8. type Address struct {
  9.   street   string
  10.   extended string
  11.   city     string
  12.   state    string
  13.   zip      string
  14. }
  16. type Name struct {
  17.   first  string
  18.   middle string
  19.   last   string
  20. }
  22. type Person struct {
  23.   name    Name
  24.   age     int
  25.   address Address
  26.   phone   string
  27. }
  29. func main() {
  30.   myname := Name{"Alfred", "H", "Eigenface"}
  31.   myaddr := Address{"42 Place Rd", "Unit 2i", "Placeton", "ST", "00921"}
  32.   me := Person{myname, 24, myaddr, "000 555-0001"}
  34.   b, err := json.Marshal(me)
  36.   if err != nil {
  37.     fmt.Println(err)
  38.   }
  40.   fmt.Println(string(b)) // err is nil, but b is empty, why?
  41.   fmt.Println("\n")
  42.   fmt.Println(me) // me is as expected, full of data
  43. }
英文:

I am using Go 1.0.3 on Mac OS X 10.8.2, and I am experimenting with the json package, trying to marshal a struct to json, but I keep getting an empty {} json object.

The err value is nil, so nothing is wrong according to the json.Marshal function, and the struct is correct. Why is this happening?

  1. package main
  3. import (
  4.   "encoding/json"
  5.   "fmt"
  6. )
  8. type Address struct {
  9.   street string
  10.   extended string
  11.   city string
  12.   state string
  13.   zip string
  14. }
  16. type Name struct {
  17.   first string
  18.   middle string
  19.   last string
  20. }
  22. type Person struct {
  23.   name Name
  24.   age int
  25.   address Address
  26.   phone string
  27. }
  29. func main() {
  30.   myname := Name{"Alfred", "H", "Eigenface"}
  31.   myaddr := Address{"42 Place Rd", "Unit 2i", "Placeton", "ST", "00921"}
  32.   me := Person{myname, 24, myaddr, "000 555-0001"}
  34.   b, err := json.Marshal(me)
  36.   if err != nil {
  37.     fmt.Println(err)
  38.   }
  40.   fmt.Println(string(b))    // err is nil, but b is empty, why?
  41.   fmt.Println("\n")
  42.   fmt.Println(me)           // me is as expected, full of data
  43. }

答案1

得分: 44

你必须将你想要进行编组的字段设置为公共的。
像这样：

  1. type Address struct {
  2.   Street string
  3.   Extended string
  4.   City string
  5.   State string
  6.   Zip string
  7. }

err 是 nil，因为所有的导出字段（在这种情况下没有）都被正确地编组了。

工作示例：https://play.golang.org/p/9NH9Bog8_C6

查看文档：http://godoc.org/encoding/json/#Marshal

英文:

You have to make the fields that you want to marshal public.
Like this:

  1. type Address struct {
  2.   Street string
  3.   Extended string
  4.   City string
  5.   State string
  6.   Zip string
  7. }

err is nil because all the exported fields, in this case there are none, were marshalled correctly.

Working example: https://play.golang.org/p/9NH9Bog8_C6

Check out the docs http://godoc.org/encoding/json/#Marshal

答案2

得分: 5

请注意，您还可以通过以下方式来操作生成的JSON中字段的名称：

  1. type Name struct {
  2.   First string `json:"firstname"`
  3.   Middle string `json:"middlename"`
  4.   Last string `json:"lastname"` 
  5. }
英文:

Note that you can also manipulate what the name of the fields in the generated JSON are by doing the following:

  1. type Name struct {
  2.   First string `json:"firstname"`
  3.   Middle string `json:"middlename"`
  4.   Last string `json:"lastname"` 
  5. }

答案3

得分: 1

JSON库无法查看结构体中的字段，除非它们是公开的。在你的情况下，字段name、age、address和phone不是公开的（以小写字母开头）。在Go语言中，变量/函数是公开的，当它们以大写字母开头时。所以为了使它工作，你的结构体需要像这样：

  1. type Person struct {
  2.   Name Name
  3.   Age int
  4.   Address Address
  5.   Phone string
  6. }
英文:

JSON library cannot view the fields in a struct unless they are public. In your case,

  1. type Person struct {
  2.   name Name
  3.   age int
  4.   address Address
  5.   phone string
  6. }

the fields name, age, address and phone are not public (start with a small letter). In golang, variables/functions are public, when they start with a capital letter. So for this to work, your struct needs to look something like this:

  1. type Person struct {
  2.   Name Name
  3.   Age int
  4.   Address Address
  5.   Phone string
  6. }

huangapple
  • 本文由 发表于 2013年3月17日 00:58:12
  • 转载请务必保留本文链接：https://go.coder-hub.com/15452004.html
  • go
  • json
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