DataType.UserDefinedTableType 方法 (String)

项目
04/26/2013

Returns an object that represents the specified type.

命名空间： Microsoft.SqlServer.Management.Smo
程序集： Microsoft.SqlServer.Smo（在 Microsoft.SqlServer.Smo.dll 中）

语法

声明
Public Shared Function UserDefinedTableType ( _
    type As String _
) As DataType
用法
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

参数

type
类型：System.String
A String value that specifies the type.

返回值

类型：Microsoft.SqlServer.Management.Smo.DataType
A DataType object value.

示例

The following code example shows how to create a user-defined table type.

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");
userTable.Schema = "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")
$userTable.Schema = "ExampleSchema"