JSON解析F＃

Question

R 7 “NETFramework \ V4.0 \资料\客户\ System.Runtime.Serialization.dll”

open System.Runtime.Serialization 
open System.Runtime.Serialization.Json 

[<DataContract>] 
    type geo = { 
     [<field: DataMember(Name = "type")>] 
     t:string 
     [<field: DataMember(Name = "coordinates")>] 
     coordinates:string 
     } 


let decode (s:string) = 
    let json = new DataContractJsonSerializer(typeof<geo>) 
    let byteArray = Encoding.UTF8.GetBytes(s) 
    let stream = new MemoryStream(byteArray) 
    json.ReadObject(stream) :?> geo 

let tw = {"type":"Point","coordinates":[-7.002648,110.449961]} 

decode tw 

这将返回 - > 结束元素从命名空间 '坐标' '' 预期。从命名空间'

中找到元素'item'如何定义DataMember坐标以便它理解？

非常感谢

Answer 1

参考System.Runtime.Serialization和System.Xml

（互动：r组成 “System.Runtime.Serialization”）

open System.IO 
open System.Runtime.Serialization.Json 
open System.Xml 
open System.Text 

/// Object to Json 
let internal json<'t> (myObj:'t) = 
     use ms = new MemoryStream() 
     (new DataContractJsonSerializer(typeof<'t>)).WriteObject(ms, myObj) 
     Encoding.Default.GetString(ms.ToArray()) 


/// Object from Json 
let internal unjson<'t> (jsonString:string) : 't = 
     use ms = new MemoryStream(ASCIIEncoding.Default.GetBytes(jsonString)) 
     let obj = (new DataContractJsonSerializer(typeof<'t>)).ReadObject(ms) 
     obj :?> 't 

Answer 2

这对我的作品

#r "System.Runtime.Serialization" 

open System.IO 
open System.Text 
open System.Runtime.Serialization 
open System.Runtime.Serialization.Json 

[<DataContract>] 
    type geo = { 
     [<field: DataMember(Name = "type")>] 
     t:string 
     [<field: DataMember(Name = "coordinates")>] 
     coordinates:float[] 
     } 


let decode (s:string) = 
    let json = new DataContractJsonSerializer(typeof<geo>) 
    let byteArray = Encoding.UTF8.GetBytes(s) 
    let stream = new MemoryStream(byteArray) 
    json.ReadObject(stream) :?> geo 

let tw = "{ 
    \"type\":\"Point\", 
    \"coordinates\":[-7.002648,110.449961] 
    }" 

let v = decode tw // val v : geo = {t = "Point"; coordinates = [|-7.002648; 110.449961|];}