Метод UserDefinedType (String, String)
Returns an object that represents the specified type with the specified schema.
Пространство имен: Microsoft.SqlServer.Management.Smo
Сборка: Microsoft.SqlServer.Smo (в Microsoft.SqlServer.Smo.dll)
Синтаксис
'Декларация
Public Shared Function UserDefinedType ( _
type As String, _
schema As String _
) As DataType
'Применение
Dim type As String
Dim schema As String
Dim returnValue As DataType
returnValue = DataType.UserDefinedType(type, _
schema)
public static DataType UserDefinedType(
string type,
string schema
)
public:
static DataType^ UserDefinedType(
String^ type,
String^ schema
)
static member UserDefinedType :
type:string *
schema:string -> DataType
public static function UserDefinedType(
type : String,
schema : String
) : DataType
Параметры
- type
Тип: System. . :: . .String
A String value that specifies the type.
- schema
Тип: System. . :: . .String
A String value that specifies the schema.
Возвращаемое значение
Тип: Microsoft.SqlServer.Management.Smo. . :: . .DataType
A DataType object value.
Примеры
The following code example shows how to create a user-defined type.
C#
Server srv = new Server("(local)");
Database db = srv.Databases["AdventureWorks2008R2"];
Schema schema1 = new Schema(db, "ExampleSchema");
schema1.Create();
UserDefinedType udt = new UserDefinedTableType(db, "udt", "ExampleSchema");
udt.Create();
DataType userType = new DataType(SqlDataType.UserDefinedType, "udt", "ExampleSchema");
Powershell
$srv = new-object Microsoft.SqlServer.Management.Smo.Server("(local)")
$db = $srv.Databases.Item("AdventureWorks2008R2")
$schema1 = new-object Microsoft.SqlServer.Management.Smo.Schema($db, "ExampleSchema")
$schema1.Create()
$udt = new-object Microsoft.SqlServer.Management.Smo.UserDefinedType($db, "udt", "ExampleSchema")
$userTable = new-object Microsoft.SqlServer.Management.Smo.DataType([Microsoft.SqlServer.Management.Smo.SqlDataType]::UserDefinedType, "udt", "ExampleSchema")
См. также