只要我有键值对解组是非常简单的,但是我将如何以不同的顺序解组不同类型的数组？单个元素是明确定义和已知的,但顺序不是.
我无法想出一个美丽的解决方案.
我会尝试错误的所有元素吗？
是否有某种联合类型可以为我做这个？
package main
import (
"encoding/json"
"fmt"
)
var my_json string = `{
"an_array":[
"with_a string",{
"and":"some_more","different":["nested","types"]
}
]
}`
type MyInner struct {
And string
Different []string
}
type MyJSON struct {
An_array []json.RawMessage
}
func main() {
var my_json_test MyJSON
e := json.Unmarshal([]byte(my_json),&my_json_test)
if e != nil {
fmt.Println(e)
} else {
for index,value := range my_json_test.An_array {
fmt.Println("index: ",index)
fmt.Println("value: ",string(value))
}
var my_inner MyInner
err := json.Unmarshal(my_json_test.An_array[1],&my_inner)
if err != nil {
fmt.Println(err)
} else {
fmt.Println("inner structure: ",my_inner)
}
}
}
解决方法
Go官方博客有一篇关于
encoding/json：
JSON and GO的很好的文章.可以将“任意数据”解码为接口{},并使用类型断言来动态地确定类型.
package main
import (
"encoding/json"
"fmt"
)
var my_json string = `{
"an_array":[
"with_a string",{
"and":"some_more","types"]
}
]
}`
func WTHisThisJSON(f interface{}) {
switch vf := f.(type) {
case map[string]interface{}:
fmt.Println("is a map:")
for k,v := range vf {
switch vv := v.(type) {
case string:
fmt.Printf("%v: is string - %q\n",k,vv)
case int:
fmt.Printf("%v: is int - %q\n",vv)
default:
fmt.Printf("%v: ",k)
WTHisThisJSON(v)
}
}
case []interface{}:
fmt.Println("is an array:")
for k,k)
WTHisThisJSON(v)
}
}
}
}
func main() {
fmt.Println("JSON:\n",my_json,"\n")
var f interface{}
err := json.Unmarshal([]byte(my_json),&f)
if err != nil {
fmt.Println(err)
} else {
fmt.Printf("JSON: ")
WTHisThisJSON(f)
}
}
输出如下：
JSON:
{
"an_array":[
"with_a string","types"]
}
]
}
JSON: is a map:
an_array: is an array:
0: is string - "with_a string"
1: is a map:
and: is string - "some_more"
different: is an array:
0: is string - "nested"
1: is string - "types"
它还不完整,但显示它将如何工作.
