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将 json 解组为反射结构(续)

[英]Unmarshal json to reflected struct (continued)

I want to write a gin middleware handler which gets data from c.Request.FormValue("data") , unmarshalls it into a structure (structures are rather different) and sets a variable in context ( c.Set("Data",newP) ).我想编写一个 gin 中间件处理程序,它从c.Request.FormValue("data")获取数据,将其解组为一个结构(结构相当不同)并在上下文中设置一个变量( c.Set("Data",newP) )。 So I searched and wrote this:所以我搜索并写了这个:

package middleware

import (
    "reflect"
    "fmt"
    "github.com/gin-gonic/gin"
    "encoding/json"
)

//https://semaphoreci.com/community/tutorials/test-driven-development-of-go-web-applications-with-gin
//https://github.com/gin-gonic/gin/issues/420
func Data(t reflect.Type) gin.HandlerFunc {
    return func(c *gin.Context) {
        //https://stackoverflow.com/a/7855298/5257901
        //https://stackoverflow.com/a/45680060/5257901
        //t := reflect.TypeOf(orig)
        v := reflect.New(t.Elem())
        // reflected pointer
        newP := v.Interface()

        data:=c.Request.FormValue("data")
        fmt.Printf("%s data:%s\n",c.Request.URL.Path,data)
        if err:=json.Unmarshal([]byte(data),newP); err!=nil{
            fmt.Printf("%s data unmarshall %s, data(in quotes):\"%s\"",c.Request.URL.Path,err,data)
            c.Abort()
            return
        }
        ustr, _:=json.Marshal(newP)
        fmt.Printf("%s unmarshalled:%s\n",c.Request.URL.Path,ustr)
        c.Set("Data",newP)
        c.Next()
    }
}

and I use it like this:我像这样使用它:

func InitHandle(R *gin.Engine) {
    Plan := R.Group("/Plan")
    Plan.POST("/clickCreate",middleware.Data(reflect.TypeOf(new(tls.PlanTabel))), clickCreatePlanHandle)
}

and

var data = *(c.MustGet("Data").(*tls.PlanTabel))

which is rather heavy and ugly.这是相当沉重和丑陋的。 I want to我想

middleware.Data(tls.PlanTabel{})

and

var data = c.MustGet("Data").(tls.PlanTabel)

In other words , omitting gin, I want a closure that eats i interface{} and returns a function (data string) (o interface{})换句话说,省略 gin,我想要一个吃i interface{}并返回一个函数(data string) (o interface{})的闭包

func Data(i interface{}) (func (string) (interface{})) {
    //some reflect magic goes here
    //extract the structure type from interface{} :

    //gets a reflect type pointer to it, like
    //t := reflect.TypeOf(orig)
    return func(data string) (o interface{}) {
        //new reflected structure (pointer?)
        v := reflect.New(t.Elem())
        //interface to it
        newP := v.Interface()
        //unmarshal
        json.Unmarshal([]byte(data),newP);
        //get the structure from the pointer back

        //returns interface to the structure

        //reflect magic ends
    }
}

The code in the question is close.问题中的代码很接近。

Try the following function.试试下面的功能。 The resulting function returns the same type as the type of i :结果函数返回与i的类型相同的类型:

func Data(i interface{}) func(string) (interface{}, error) {
    return func(data string) (interface{}, error) {
        v := reflect.New(reflect.TypeOf(i))
        err := json.Unmarshal([]byte(data), v.Interface())
        return v.Elem().Interface(), err
    }
}

Example use:使用示例:

type Test struct {
    A string
    B string
}

f := Data((*Test)(nil))
v, err := f(`{"A": "hello", "B": "world"}`)
if err != nil {
    log.Fatal(err)
}
fmt.Printf("%#v\n", v) // v is a *Test

f = Data("")
v, err = f(`"Hello"`)
if err != nil {
    log.Fatal(err)
}
fmt.Printf("%#v\n", v)  // v is a string

If you want to return a struct value, then pass a struct value as the argument to Data:如果要返回结构值,则将结构值作为参数传递给 Data:

f = Data(Test{})
v, err = f(`{"A": "hello", "B": "world"}`)
if err != nil {
    log.Fatal(err)
}
fmt.Printf("%#v\n", v) // v is a Test

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