SerializeConvertFromJsonString(String, Type, JsonSerializerOptions) 方法

public static Object FromJsonString(
	this string json,
	Type type,
	JsonSerializerOptions jsonSerializerOptions = null
)

<ExtensionAttribute>
Public Shared Function FromJsonString ( 
	json As String,
	type As Type,
	Optional jsonSerializerOptions As JsonSerializerOptions = Nothing
) As Object

[<ExtensionAttribute>]
static member FromJsonString : 
        json : string * 
        type : Type * 
        ?jsonSerializerOptions : JsonSerializerOptions 
(* Defaults:
        let _jsonSerializerOptions = defaultArg jsonSerializerOptions null
*)
-> Object 

参数

json  String

[缺少 "M:TouchSocket.Core.SerializeConvert.FromJsonString(System.String,System.Type,System.Text.Json.JsonSerializerOptions)" 的 <param name="json"/> 文档]

type  Type

[缺少 "M:TouchSocket.Core.SerializeConvert.FromJsonString(System.String,System.Type,System.Text.Json.JsonSerializerOptions)" 的 <param name="type"/> 文档]

jsonSerializerOptions  JsonSerializerOptions  (Optional)

[缺少 "M:TouchSocket.Core.SerializeConvert.FromJsonString(System.String,System.Type,System.Text.Json.JsonSerializerOptions)" 的 <param name="jsonSerializerOptions"/> 文档]

返回值

[缺少 "M:TouchSocket.Core.SerializeConvert.FromJsonString(System.String,System.Type,System.Text.Json.JsonSerializerOptions)" 的 <returns> 文档]

备注

引用