UserDefinedTableType 메서드 (String)
Returns an object that represents the specified type.
네임스페이스: Microsoft.SqlServer.Management.Smo
어셈블리: Microsoft.SqlServer.Smo(Microsoft.SqlServer.Smo.dll)
구문
‘선언
Public Shared Function UserDefinedTableType ( _
type As String _
) As DataType
‘사용 방법
Dim type As String
Dim returnValue As DataType
returnValue = DataType.UserDefinedTableType(type)
public static DataType UserDefinedTableType(
string type
)
public:
static DataType^ UserDefinedTableType(
String^ type
)
static member UserDefinedTableType :
type:string -> DataType
public static function UserDefinedTableType(
type : String
) : DataType
매개 변수
- type
유형: System. . :: . .String
A String value that specifies the type.
반환 값
유형: Microsoft.SqlServer.Management.Smo. . :: . .DataType
A DataType object value.
예
The following code example shows how to create a user-defined table type.
C#
Server srv = new Server("(local)");
Database db = srv.Databases["AdventureWorks2008R2"];
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");
userTable.Schema = "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()
$udtt = new-object Microsoft.SqlServer.Management.Smo.UserDefinedTableType($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")
$userTable.Schema = "ExampleSchema"