获取 JSON 动态键名作为字符串？

huangapple
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英文:

Getting json dynamic key name as string?

问题

例如：

  1. {"id":
  2.     {"12345678901234":
  3.        {"Account":"asdf",
  4.        "Password":"qwerty"
  5.        "LastSeen":"1397621470",
  6.        }
  7.     }
  8. }

我一直在尝试编写一个程序，需要将id作为字符串获取，然后稍后使用它来检查LastSeen中的时间。我尝试使用simplejson和jsonq，但仍然无法弄清楚如何做到这一点。

英文:

For example:

  1. {"id":
  2.     {"12345678901234":
  3.        {"Account":"asdf",
  4.        "Password":"qwerty"
  5.        "LastSeen":"1397621470",
  6.        }
  7.     }
  8. }

A program I've been trying to make needs to get the id as a string and then later use it to check the time in LastSeen.
I've tried using simplejson and jsonq,but still cant figure out how to do that.

答案1

得分: 3

你可以使用RawMessage来简化代码（play with it) :

  1. package main
  3. import (
  4. 	"encoding/json"
  5. 	"fmt"
  6. )
  8. var data []byte = []byte(`{"id": {"12345678901234": {"Account":"asdf", "Password":"qwerty", "LastSeen":"1397621470"}}}`)
  10. type Message struct {
  11. 	Id   string
  12. 	Info struct {
  13. 		Account  string
  14. 		Password string
  15. 		LastSeen string
  16. 	}
  17. }
  19. func main() {
  20. 	var (
  21. 		tmpmsg struct {
  22. 			Data map[string]json.RawMessage `json:"id"`
  23. 		}
  24. 		msg Message
  25. 	)
  26. 	if err := json.Unmarshal(data, &tmpmsg); err != nil {
  27. 		panic(err) //you probably wanna use or something instead
  28. 	}
  30. 	for id, raw := range tmpmsg.Data {
  31. 		msg.Id = id
  32. 		if err := json.Unmarshal(raw, &msg.Info); err != nil {
  33. 			panic(err)
  34. 		}
  35. 	}
  36. 	fmt.Printf("%+v\n", msg)
  38. }
英文:

You can use RawMessage and make it much simpiler (play with it) :

  1. package main
  3. import (
  4. 	"encoding/json"
  5. 	"fmt"
  6. )
  8. var data []byte = []byte(`{"id": {"12345678901234": {"Account":"asdf", "Password":"qwerty", "LastSeen":"1397621470"}}}`)
  10. type Message struct {
  11. 	Id   string
  12. 	Info struct {
  13. 		Account  string
  14. 		Password string
  15. 		LastSeen string
  16. 	}
  17. }
  19. func main() {
  20. 	var (
  21. 		tmpmsg struct {
  22. 			Data map[string]json.RawMessage `json:"id"`
  23. 		}
  24. 		msg Message
  25. 	)
  26. 	if err := json.Unmarshal(data, &tmpmsg); err != nil {
  27. 		panic(err) //you probably wanna use or something instead
  28. 	}
  30. 	for id, raw := range tmpmsg.Data {
  31. 		msg.Id = id
  32. 		if err := json.Unmarshal(raw, &msg.Info); err != nil {
  33. 			panic(err)
  34. 		}
  35. 	}
  36. 	fmt.Printf("%+v\n", msg)
  38. }

答案2

得分: 1

在Golang博客上关于JSON的文章中，可以使用encoding/json包来实现。我创建了一个小程序来实现这个功能，代码如下：

  1. package main
  3. import (
  4.     "encoding/json"
  5.     "fmt"
  6. )
  8. var data []byte = []byte(`{"id": {"12345678901234": {"Account":"asdf", "Password":"qwerty", "LastSeen":"1397621470"}}}`)
  10. type Message struct {
  11.     id string
  12.     LastSeen int64
  13. } 
  15. var m Message
  17. func main() {
  18.     var i interface {}
  19.     err := json.Unmarshal(data, &i)
  20.     if err != nil {
  21.         println("Error decoding data")
  22.         fmt.Printf("%s", err.Error())
  23.         return
  24.     }
  25.     m := i.(map[string]interface{})
  27.     for k, v := range m {
  28.         println(k)
  29.         im := v.(map[string]interface{})
  30.         for ik, iv := range im {
  31.             println("\t", ik)
  32.             jm := iv.(map[string]interface{})
  33.             for jk, jv := range jm {
  34.                 println("\t\t", jk, ": ", jv.(string))
  35.             }
  36.         }
  37.     }
  38. }

如果这段代码在Go的最佳实践方面存在问题，我向你道歉，因为我对这门语言还不熟悉。我知道其中有一些部分并不是完全必要的，比如Message类型的定义，但至少在你的数据上是可以工作的。

英文:

Looking at the Golang blog post on JSON <a href="http://blog.golang.org/json-and-go">here</a> it can be done using the encoding/json package. I created a small program to do this as follows:

  1. package main
  3. import (
  4.     &quot;encoding/json&quot;
  5.     &quot;fmt&quot;
  6. )
  8. var data []byte = []byte(`{&quot;id&quot;: {&quot;12345678901234&quot;: {&quot;Account&quot;:&quot;asdf&quot;, &quot;Password&quot;:&quot;qwerty&quot;, &quot;LastSeen&quot;:&quot;1397621470&quot;}}}`)
  10. type Message struct {
  11.     id string
  12.     LastSeen int64
  13. } 
  15. var m Message
  17. func main() {
  18.     var i interface {}
  19.     err := json.Unmarshal(data, &amp;i)
  20.     if err != nil {
  21.         println(&quot;Error decoding data&quot;)
  22.         fmt.Printf(&quot;%s&quot;, err.Error())
  23.         return
  24.     }
  25.     m := i.(map[string]interface{})
  27.     for k, v := range m {
  28.         println(k)
  29.         im := v.(map[string]interface{})
  30.         for ik, iv := range im {
  31.             println(&quot;\t&quot;, ik)
  32.             jm := iv.(map[string]interface{})
  33.             for jk, jv := range jm {
  34.                 println(&quot;\t\t&quot;, jk, &quot;: &quot;, jv.(string))
  35.             }
  36.         }
  37.     }
  38. }

I apologise if this is poor in terms of Go best practices and such, I am new to the language. And I know that some elements of this aren't entirely necessary like the Message type definition but this works, at least on your data.

huangapple
  • 本文由 发表于 2014年4月25日 18:33:09
  • 转载请务必保留本文链接：https://go.coder-hub.com/23290618.html
  • go
  • json
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