Interface ILambdaJsonSerializer
The ILambdaJsonSerializer interface extends ILambdaSerializer interface by adding
the ability to deserialize a JSON document to an explicitly specified type instead of relying on a
generic type parameter.
Namespace: LambdaSharp.Serialization
Assembly: LambdaSharp.dll
Syntax
public interface ILambdaJsonSerializer : ILambdaSerializerMethods
Deserialize(Stream, Type)
The Deserialize(Stream, Type) method deserializes the JSON object from a Stream.
Declaration
object Deserialize(Stream stream, Type type)Parameters
| Type | Name | Description |
|---|---|---|
| System.IO.Stream | stream | Stream to deserialize. |
| System.Type | type | The type to instantiate. |
Returns
| Type | Description |
|---|---|
| System.Object | Deserialized instance. |
Deserialize(String, Type)
The Deserialize(String, Type) method deserializes the JSON object from a string.
Declaration
virtual object Deserialize(string json, Type type)Parameters
| Type | Name | Description |
|---|---|---|
| System.String | json | String to deserialize. |
| System.Type | type | The type to instantiate. |
Returns
| Type | Description |
|---|---|
| System.Object | Deserialized instance. |
Deserialize<T>(String)
The Deserialize<T>(String) method deserializes the JSON object from a string.
Declaration
virtual T Deserialize<T>(string json)Parameters
| Type | Name | Description |
|---|---|---|
| System.String | json | The |
Returns
| Type | Description |
|---|---|
| T | Deserialized instance. |
Type Parameters
| Name | Description |
|---|---|
| T | The deserialization target type. |
Serialize<T>(T)
The Serialize<T>(T) interface method serializes
an instance to a JSON string.
Declaration
virtual string Serialize<T>(T value)Parameters
| Type | Name | Description |
|---|---|---|
| T | value | The instance to serialize. |
Returns
| Type | Description |
|---|---|
| System.String | Serialized JSON |
Type Parameters
| Name | Description |
|---|---|
| T |