DataType.UserDefinedTableType Método (String, String)
Returns an object that represents the specified type with the specified schema.
Espacio de nombres: Microsoft.SqlServer.Management.Smo
Ensamblado: Microsoft.SqlServer.Smo (en Microsoft.SqlServer.Smo.dll)
Sintaxis
'Declaración
Public Shared Function UserDefinedTableType ( _
type As String, _
schema As String _
) As DataType
'Uso
Dim type As String
Dim schema As String
Dim returnValue As DataType
returnValue = DataType.UserDefinedTableType(type, _
schema)
public static DataType UserDefinedTableType(
string type,
string schema
)
public:
static DataType^ UserDefinedTableType(
String^ type,
String^ schema
)
static member UserDefinedTableType :
type:string *
schema:string -> DataType
public static function UserDefinedTableType(
type : String,
schema : String
) : DataType
Parámetros
- type
Tipo: System.String
A String value that specifies the type.
- schema
Tipo: System.String
A String value that specifies the schema.
Valor devuelto
Tipo: Microsoft.SqlServer.Management.Smo.DataType
A DataType object value.
Ejemplos
The following code example shows how to create a user-defined table type.
C#
Server srv = new Server("(local)");
Database db = srv.Databases["AdventureWorks2012"];
Schema schema1 = new Schema(db, "ExampleSchema");
schema1.Create();
UserDefinedTableType udtt = new UserDefinedTableType(db, "udtt", "ExampleSchema");
Column c = new Column(udtt, "Column 1", DataType.Int);
udtt.Columns.Add(c);
udtt.Create();
DataType userTable = new DataType(SqlDataType.UserDefinedTableType, "udtt", "ExampleSchema");
Powershell
$srv = new-object Microsoft.SqlServer.Management.Smo.Server("(local)")
$db = $srv.Databases.Item("AdventureWorks2012")
$schema1 = new-object Microsoft.SqlServer.Management.Smo.Schema($db, "ExampleSchema")
$schema1.Create()
$udtt = new-object Microsoft.SqlServer.Management.Smo.UserDefinedTableType($db, "udtt", "ExampleSchema")
$c = new-object Microsoft.SqlServer.Management.Smo.Column($udtt, "Column 1", [Microsoft.SqlServer.Management.Smo.DataType]::Int)
$udtt.Columns.Add($c)
$udtt.Create()
$userTable = new-object Microsoft.SqlServer.Management.Smo.DataType([Microsoft.SqlServer.Management.Smo.SqlDataType]::UserDefinedTableType, "udtt", "ExampleSchema")
Vea también
Referencia
Sobrecarga de UserDefinedTableType
Espacio de nombres Microsoft.SqlServer.Management.Smo