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Propriedade do RDL FileGroup

Obtém ou define o grupo de arquivos em que a tabela está armazenada.

Namespace:  Microsoft.SqlServer.Management.Smo
Assembly:  Microsoft.SqlServer.Smo (em Microsoft.SqlServer.Smo.dll)

Sintaxe

'Declaração
<SfcPropertyAttribute(SfcPropertyFlags.None Or SfcPropertyFlags.ReadOnlyAfterCreation Or SfcPropertyFlags.Standalone)> _
Public Property FileGroup As String
    Get
    Set
'Uso
Dim instance As Table
Dim value As String

value = instance.FileGroup

instance.FileGroup = value

[SfcPropertyAttribute(SfcPropertyFlags.None|SfcPropertyFlags.ReadOnlyAfterCreation|SfcPropertyFlags.Standalone)]
public string FileGroup { get; set; }

[SfcPropertyAttribute(SfcPropertyFlags::None|SfcPropertyFlags::ReadOnlyAfterCreation|SfcPropertyFlags::Standalone)]
public:
property String^ FileGroup {
    String^ get ();
    void set (String^ value);
}

[<SfcPropertyAttribute(SfcPropertyFlags.None|SfcPropertyFlags.ReadOnlyAfterCreation|SfcPropertyFlags.Standalone)>]
member FileGroup : string with get, set

function get FileGroup () : String
function set FileGroup (value : String)

Valor da propriedade

Tipo: System. . :: . .String
Um valor String que especifica o grupo de arquivos no qual a tabela está armazenada.

Comentários

The FileGroup property is set prior to the creation of the table. After the table is created, the property is read-only.

Exemplos

The following code example shows how to create a new table and display the time and date that it was created on the console.

C#

Server srv = new Server("(local)");
Database db = srv.Databases["AdventureWorks2008R2"];

Table tb = new Table(db, "Test Table");
Column col1 = new Column(tb, "Name", DataType.NChar(50));
Column col2 = new Column(tb, "ID", DataType.Int);

tb.Columns.Add(col1); 
tb.Columns.Add(col2); 
tb.Create();

Console.WriteLine("The table is part of the " + tb.FileGroup.ToString() + " file group.");

Powershell

$srv = new-Object Microsoft.SqlServer.Management.Smo.Server("(local)")
$db = New-Object Microsoft.SqlServer.Management.Smo.Database
$db = $srv.Databases.Item("AdventureWorks2008R2")

#Create the Table
$tb = new-object Microsoft.SqlServer.Management.Smo.Table($db, "Test Table")
$col1 = new-object Microsoft.SqlServer.Management.Smo.Column($tb, "Name", [Microsoft.SqlServer.Management.Smo.DataType]::NChar(50))
$col2 = new-object Microsoft.SqlServer.Management.Smo.Column($tb, "ID", [Microsoft.SqlServer.Management.Smo.DataType]::Int)
$tb.Columns.Add($col1)
$tb.Columns.Add($col2)
$tb.Create()

Write-Host "The table is part of the" $tb.FileGroup "file group."