JsonTypeInfo.CreateJsonPropertyInfo(Type, String) 方法

public:
 System::Text::Json::Serialization::Metadata::JsonPropertyInfo ^ CreateJsonPropertyInfo(Type ^ propertyType, System::String ^ name);

[System.Diagnostics.CodeAnalysis.RequiresDynamicCode("JSON serialization and deserialization might require types that cannot be statically analyzed and might need runtime code generation. Use System.Text.Json source generation for native AOT applications.")]
[System.Diagnostics.CodeAnalysis.RequiresUnreferencedCode("JSON serialization and deserialization might require types that cannot be statically analyzed and might need runtime code generation. Use System.Text.Json source generation for native AOT applications.")]
public System.Text.Json.Serialization.Metadata.JsonPropertyInfo CreateJsonPropertyInfo(Type propertyType, string name);

public System.Text.Json.Serialization.Metadata.JsonPropertyInfo CreateJsonPropertyInfo(Type propertyType, string name);

[<System.Diagnostics.CodeAnalysis.RequiresDynamicCode("JSON serialization and deserialization might require types that cannot be statically analyzed and might need runtime code generation. Use System.Text.Json source generation for native AOT applications.")>]
[<System.Diagnostics.CodeAnalysis.RequiresUnreferencedCode("JSON serialization and deserialization might require types that cannot be statically analyzed and might need runtime code generation. Use System.Text.Json source generation for native AOT applications.")>]
member this.CreateJsonPropertyInfo : Type * string -> System.Text.Json.Serialization.Metadata.JsonPropertyInfo

member this.CreateJsonPropertyInfo : Type * string -> System.Text.Json.Serialization.Metadata.JsonPropertyInfo

Public Function CreateJsonPropertyInfo (propertyType As Type, name As String) As JsonPropertyInfo

參數

傳回

一個空白 JsonPropertyInfo 的實例。

屬性

RequiresDynamicCodeAttribute RequiresUnreferencedCodeAttribute

例外狀況