我的结构体无法转换为json。

huangapple go评论86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?

package main

import (
  "encoding/json"
  "fmt"
)

type Address struct {
  street   string
  extended string
  city     string
  state    string
  zip      string
}

type Name struct {
  first  string
  middle string
  last   string
}

type Person struct {
  name    Name
  age     int
  address Address
  phone   string
}

func main() {
  myname := Name{"Alfred", "H", "Eigenface"}
  myaddr := Address{"42 Place Rd", "Unit 2i", "Placeton", "ST", "00921"}
  me := Person{myname, 24, myaddr, "000 555-0001"}

  b, err := json.Marshal(me)

  if err != nil {
    fmt.Println(err)
  }

  fmt.Println(string(b)) // err is nil, but b is empty, why?
  fmt.Println("\n")
  fmt.Println(me) // me is as expected, full of data
}
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?

package main

import (
  "encoding/json"
  "fmt"
)

type Address struct {
  street string
  extended string
  city string
  state string
  zip string
}

type Name struct {
  first string
  middle string
  last string
}

type Person struct {
  name Name
  age int
  address Address
  phone string
}

func main() {
  myname := Name{"Alfred", "H", "Eigenface"}
  myaddr := Address{"42 Place Rd", "Unit 2i", "Placeton", "ST", "00921"}
  me := Person{myname, 24, myaddr, "000 555-0001"}

  b, err := json.Marshal(me)

  if err != nil {
    fmt.Println(err)
  }

  fmt.Println(string(b))    // err is nil, but b is empty, why?
  fmt.Println("\n")
  fmt.Println(me)           // me is as expected, full of data
}

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:

type Address struct {
  Street string
  Extended string
  City string
  State string
  Zip string
}

errnil,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:

type Address struct {
  Street string
  Extended string
  City string
  State string
  Zip string
}

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:

type Name struct {
  First string `json:"firstname"`
  Middle string `json:"middlename"`
  Last string `json:"lastname"` 
}
英文:

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

type Name struct {
  First string `json:"firstname"`
  Middle string `json:"middlename"`
  Last string `json:"lastname"` 
}

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:

type Person struct {
  Name Name
  Age int
  Address Address
  Phone string
}
英文:

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

type Person struct {
  name Name
  age int
  address Address
  phone string
}

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:

type Person struct {
  Name Name
  Age int
  Address Address
  Phone string
}

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
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