从F#调用Newtonsoft.Json的意外结果 [英] Unexpected result from call to Newtonsoft.Json from F#

问题描述

我没有从此F#代码中获得预期的结果.我希望t包含作为对JsonSchema.Parse(json)的调用的结果，但它为空.我在做什么错了?

I am not getting the expected result from this F# code. I would expect t to contain values as a result of the call to JsonSchema.Parse(json) but instead it is empty. What am I doing wrong?

open Newtonsoft.Json
open Newtonsoft.Json.Schema

let json = """{
  "Name": "Bad Boys",
  "ReleaseDate": "1995-4-7T00:00:00",
  "Genres": [
    "Action",
    "Comedy"
  ]
}"""

[<EntryPoint>]
let main argv = 
    let t  = JsonSchema.Parse(json)  
    0 // return an integer exit code

推荐答案

正如John Palmer指出的那样，JsonSchema.Parse解析JSON schema ，但是从您的问题来看，好像您想解析普通的JSON 值.使用JsonConvert.DeserializeObject:

As John Palmer points out, JsonSchema.Parse parses a JSON schema, but from your question, it looks as though you want to parse a normal JSON value. This is possible with JsonConvert.DeserializeObject:

let t = JsonConvert.DeserializeObject json

但是，DeserializeObject的签名将返回obj，因此这对访问值没有特别帮助.为此，必须将返回值强制转换为JObject:

However, the signature of DeserializeObject is to return obj, so that doesn't particularly help you access the values. In order to do so, you must cast the return value to JObject:

let t = (JsonConvert.DeserializeObject json) :?> Newtonsoft.Json.Linq.JObject
let name = t.Value<string> "Name"

Json.NET旨在利用C#的dynamic关键字，但F#并未内置该关键字的完全等效项.但是，您可以通过 FSharp.Dynamic

Json.NET is designed to take advantage of C#'s dynamic keyword, but the exact equivalent of that isn't built into F#. However, you can get a similar syntax via FSharp.Dynamic:

open EkonBenefits.FSharp.Dynamic

let t = JsonConvert.DeserializeObject json
let name = t?Name

在Name之前注意?.请记住，JSON区分大小写.

Notice the ? before Name. Keep in mind that JSON is case-sensitive.

现在，name仍然不是字符串，而是一个JValue对象，但是您可以通过在其上调用ToString()来获取字符串值，但是您也可以使用JValue的Value属性，如果该值是数字而不是字符串，则可以很方便:

Now, name still isn't a string, but rather a JValue object, but you can get the string value by calling ToString() on it, but you can also use JValue's Value property, which can be handy if the value is a number instead of a string:

let jsonWithNumber = """{ "number" : 42 }"""
let t = JsonConvert.DeserializeObject jsonWithNumber
let actual = t?number?Value
Assert.Equal(42L, actual)

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