遇到了解析未知键的嵌套 JSON 的问题。

huangapple
go评论114阅读模式
英文:

Trouble unmarshalling nested json with unknown keys

问题

我遇到了一些问题，无法将下面格式的JSON数据解组成一个结构体。JSON的结构对我来说有点混乱，对于我做的所有愚蠢的解组尝试，我表示抱歉。

  1. {
  2.   "message": {
  3.     "Server1.example.com": [
  4.       {
  5.         "application": "Apache",
  6.         "host": {
  7.           "name": "/^Server-[13456]/"
  8.         },
  9.         "owner": "User1",
  10.         "project": "Web",
  11.         "subowner": "User2"
  12.       }
  13.     ],
  14.     "Server2.example.com": [
  15.       {
  16.         "application": "Mysql",
  17.         "host": {
  18.           "name": "/^Server[23456]/"
  19.         },
  20.         "owner": "User2",
  21.         "project": "DB",
  22.         "subowner": "User3"
  23.       }
  24.     ]
  25.   },
  26.   "response_ms": 659,
  27.   "success": true
  28. }

我尝试使用以下结构体进行解组：

  1. type ServerDetails struct {
  2.   Message struct {
  3.     Hostname struct {
  4.       Details struct {
  5.         Application string `json:"application"`
  6.       } `json:"-"`
  7.     } `json:"-"`
  8.   } `json:"message"`
  9. }

在生成时，Server[0-9].example.com字段是未知的，并且会发生变化。而且，在服务器名称之后，有一个没有外部键的字段，这再次让我感到困惑。我尝试了很多组合来理解如何解组，但是失败了。

有什么有效的方法可以将JSON字段解组到结构体中吗？

英文:

I am having trouble unmarshalling a json data of the below format to a struct. The structure of the json looks a bit confusing to me, so apologies for all the dumb things I am doing to unmarshal it.

  1. {
  2.   "message": {
  3.     "Server1.example.com": [
  4.       {
  5.         "application": "Apache", 
  6.         "host": {
  7.           "name": "/^Server-[13456]/"
  8.         }, 
  9.         "owner": "User1", 
  10.         "project": "Web", 
  11.         "subowner": "User2"
  12.       }
  13.     ], 
  14.     "Server2.example.com": [
  15.       {
  16.         "application": "Mysql", 
  17.         "host": {
  18.           "name": "/^Server[23456]/"
  19.         }, 
  20.         "owner": "User2", 
  21.         "project": "DB", 
  22.         "subowner": "User3"
  23.       }
  24.     ] 
  25.   }, 
  26.   "response_ms": 659, 
  27.   "success": true
  28. }

I am trying to unmarshal it using the following struct.

  1. type ServerDetails struct  {
  2.   Message  struct{
  3.     Hostname struct{
  4.       Details struct{
  5.         Application string `json:"application"`
  6.         }`json:"-"`
  7.        }`json:"-"`
  8.      }`json:"message"`
  9. }

The fields Server[0-9].example.com will be unknown at the time of generating, and will change, and there is this field

  1.       {
  2.     "application": "Apache", 
  3.     "host": {
  4.       "name": "/^Server-[13456]/"
  5.     },

just after the server name that doesn't have a key outside, which again looks confusing to me. I tried a good number of combinations to understand how this could be unmarshalled, but I failed.

What is a working approach to get the json fields unmarshal into a struct?

答案1

得分: 3

你可以使用map[string]ServerStruct来满足你的需求。

你的结构体可以像这样定义：

  1. type YourStruct struct {
  2.     Success    bool
  3.     ResponseMS int
  4.     Servers    map[string]*ServerStruct
  5. }
  7. type ServerStruct struct {
  8.     Application string
  9.     Owner       string
  10.     // 其他字段...
  11. }

通过添加一些额外的JSON标签，你就可以解析你的JSON数据了。

英文:

You could include a map[string]ServerStruct to fullfill your requirements.

your struct could look like this:

  1. type YourStruct struct {
  2.     Success bool
  3.     ResponseMS int
  4.     Servers map[string]*ServerStruct
  5. }
  7. type ServerStruct struct {
  8.     Application string
  9.     Owner string
  10.     [...]
  11. }

With some additional json tags, you will be able to parse your json.

答案2

得分: 0

你的JSON在第二个"]"后面有多余的逗号，不符合规范。一旦你修正了JSON，你可以使用优秀的https://mholt.github.io/json-to-go/来构建以下Go结构体：

  1. type AutoGenerated struct {
  2.     Message struct {
  3.         Server1ExampleCom []struct {
  4.             Application string `json:"application"`
  5.             Host struct {
  6.                 Name string `json:"name"`
  7.             } `json:"host"`
  8.             Owner string `json:"owner"`
  9.             Project string `json:"project"`
  10.             Subowner string `json:"subowner"`
  11.         } `json:"Server1.example.com"`
  12.         Server2ExampleCom []struct {
  13.             Application string `json:"application"`
  14.             Host struct {
  15.                 Name string `json:"name"`
  16.             } `json:"host"`
  17.             Owner string `json:"owner"`
  18.             Project string `json:"project"`
  19.             Subowner string `json:"subowner"`
  20.         } `json:"Server2.example.com"`
  21.     } `json:"message"`
  22.     ResponseMs int `json:"response_ms"`
  23.     Success bool `json:"success"`
  24. }
英文:

You JSON is not valid with superfluous comma after the second ]
Once you correct the JSON, you can use the excellent https://mholt.github.io/json-to-go/ to build the following Go struct

  1. type AutoGenerated struct {
  2. 	Message struct {
  3. 		Server1ExampleCom []struct {
  4. 			Application string `json:"application"`
  5. 			Host struct {
  6. 				Name string `json:"name"`
  7. 			} `json:"host"`
  8. 			Owner string `json:"owner"`
  9. 			Project string `json:"project"`
  10. 			Subowner string `json:"subowner"`
  11. 		} `json:"Server1.example.com"`
  12. 		Server2ExampleCom []struct {
  13. 			Application string `json:"application"`
  14. 			Host struct {
  15. 				Name string `json:"name"`
  16. 			} `json:"host"`
  17. 			Owner string `json:"owner"`
  18. 			Project string `json:"project"`
  19. 			Subowner string `json:"subowner"`
  20. 		} `json:"Server2.example.com"`
  21. 	} `json:"message"`
  22. 	ResponseMs int `json:"response_ms"`
  23. 	Success bool `json:"success"`
  24. }

huangapple
  • 本文由 发表于 2016年11月9日 19:03:19
  • 转载请务必保留本文链接：https://go.coder-hub.com/40505410.html
  • go
  • json
  • unmarshalling
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定