英文:
How does Golang accept uncertain values
问题
后端返回的值是不固定的,有时候是这样的:
{"application": {"instance": [{"instanceId": "v1"}, {"instanceId": "v2"}]}}
有时候是这样的:
{"application": {"instance": {"instanceId": "v"}}}
我应该如何提取对应的instanceId值?
package main
import (
"encoding/json"
"fmt"
)
type Application struct {
Application struct {
Instance json.RawMessage `json:"instance"`
} `json:"application"`
}
func main() {
a := `{"application": {"instance": {"instanceId": "v"}}}`
//a := `{"application": {"instance": [{"instanceId": "v1"}, {"instanceId": "v2"}]}}`
var p Application
errJson := json.Unmarshal([]byte(a), &p)
if errJson != nil {
fmt.Printf("errJson")
}
fmt.Printf("type:%T", p.Application.Instance)
}
英文:
Back-end return values are not fixed, sometimes:
{"application": {"instance": [{"instanceId": "v1"}, {"instanceId": "v2"}]}}
or sometimes:
{"application": {"instance": {"instanceId": "v"}}}
how should I take out the corresponding instanceId value?
package main
import (
"encoding/json"
"fmt"
)
type Application struct {
Application struct {
Instance json.RawMessage `json:"instance"`
} `json:"application"`
}
func main() {
a := `{"application": {"instance": {"instanceId": "v"}}}`
//a := `{"application": {"instance": [{"instanceId": "v1"}, {"instanceId": "v2"}]}} `
var p Application
errJson := json.Unmarshal([]byte(a), &p)
if errJson != nil {
fmt.Printf("errJson")
}
fmt.Printf("type:%T", p.Application.Instance)
}
答案1
得分: 2
由于两个值类型冲突(一个是结构体,另一个是结构体切片),将其封装为单个类型变得混乱,即使使用interface{}
等通用解决方案也无法解决。
最简单的解决方案是定义两个不同的类型,并将其解组为其中一个以查看哪个“有效”:
func unmarsh(body []byte) (*type1, *type2, error) {
var (
t1 type1
t2 type2
)
err := json.Unmarshal(body, &t1)
if err == nil {
return &t1, nil, nil
}
err = json.Unmarshal(body, &t2)
if err == nil {
return nil, &t2, nil
}
return nil, nil, err
}
在你的示例中,这两个类型将是:
type type1 struct {
Application struct {
Instance []struct {
InstanceID string `json:"instanceId"`
} `json:"instance"`
} `json:"application"`
}
type type2 struct {
Application struct {
Instance struct {
InstanceID string `json:"instanceId"`
} `json:"instance"`
} `json:"application"`
}
工作示例:
https://play.golang.org/p/Kma32gWfghb
英文:
Since the 2 value types clash (one a struct another a slice of structs) it gets messy to encapsulate this into a single type even using catch-all solutions like interface{}
.
The simplest solution is present two distinct types and marshal into either to see which "works":
func unmarsh(body []byte) (*type1, *type2, error) {
var (
t1 type1
t2 type2
)
err := json.Unmarshal(body, &t1)
if err == nil {
return &t1, nil, nil
}
err = json.Unmarshal(body, &t2)
if err == nil {
return nil, &t2, nil
}
return nil, nil, err
}
and in your example the two types would be:
type type1 struct {
Application struct {
Instance []struct {
InstanceID string `json:"instanceId"`
} `json:"instance"`
} `json:"application"`
}
type type2 struct {
Application struct {
Instance struct {
InstanceID string `json:"instanceId"`
} `json:"instance"`
} `json:"application"`
}
Working example:
https://play.golang.org/p/Kma32gWfghb
答案2
得分: 2
一个更简洁的解决方案是使用自定义的解组器(unmarshaler):
type Instances []Instance
func (i *Instances) UnmarshalJSON(in []byte) error {
if len(in) > 0 && in[0] == '[' {
var a []Instance
if err := json.Unmarshal(in, &a); err != nil {
return err
}
*i = a
return nil
}
var s Instance
if err := json.Unmarshal(in, &s); err != nil {
return err
}
*i = []Instance{s}
return nil
}
这将把一个对象解组成一个长度为1的切片。
@mkopriva 提供了一个更紧凑的解决方案:
func (i *Instances) UnmarshalJSON(in []byte) error {
if len(in) > 0 && in[0] == '[' {
return json.Unmarshal(in, (*[]Instance)(i))
}
*i = Instances{{}}
return json.Unmarshal(in, &(*i)[0])
}
英文:
A cleaner solution would be a custom unmarshaler:
type Instances []Instance
func (i *Instances) UnmarshalJSON(in []byte) error {
if len(in)>0 && in[0]=='[' {
var a []Instance
if err:=json.Unmarshal(in,&a); err!=nil {
return err
}
*i=a
return nil
}
var s Instance
if err:=json.Unmarshal(in,&s) ; err!=nil {
return err
}
*i=[]Instance{s}
return nil
}
This would unmarshal an object into a slice of 1.
A more compact solution is provided by @mkopriva:
func (i *Instances) UnmarshalJSON(in []byte) error {
if len(in) > 0 && in[0] == '[' {
return json.Unmarshal(in, (*[]Instance)(i))
}
*i = Instances{{}}
return json.Unmarshal(in, &(*i)[0])
}
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库,让每个人都能够通过互相帮助和分享经验来进步。
评论