Costruttore DataType (SqlDataType, Int32, Int32)
Initializes a new instance of the DataType class based on a specified SQL Server data type with the specified precision and scale.
Spazio dei nomi Microsoft.SqlServer.Management.Smo
Assembly: Microsoft.SqlServer.Smo (in Microsoft.SqlServer.Smo.dll)
Sintassi
'Dichiarazione
Public Sub New ( _
sqlDataType As SqlDataType, _
precision As Integer, _
scale As Integer _
)
'Utilizzo
Dim sqlDataType As SqlDataType
Dim precision As Integer
Dim scale As Integer
Dim instance As New DataType(sqlDataType, _
precision, scale)
public DataType(
SqlDataType sqlDataType,
int precision,
int scale
)
public:
DataType(
SqlDataType sqlDataType,
int precision,
int scale
)
new :
sqlDataType:SqlDataType *
precision:int *
scale:int -> DataType
public function DataType(
sqlDataType : SqlDataType,
precision : int,
scale : int
)
Parametri
- sqlDataType
Tipo: Microsoft.SqlServer.Management.Smo. . :: . .SqlDataType
A SqlDataType object value that specifies the SQL Server data type.
- precision
Tipo: System. . :: . .Int32
An Int32 value that specifies the precision for numeric SQL Server data types.
- scale
Tipo: System. . :: . .Int32
An Int32 value that specifies the scale for numeric SQL Server data types.
Esempi
Visual Basic
'Declare and create a DataType object variable.
Dim dt As DataType
dt = New DataType(SqlDataType.Decimal, 3, 3)
PowerShell
$dt = new-object Microsoft.SqlServer.Management.Smo.DataType([Microsoft.SqlServer.Management.Smo.SqlDataType]::Decimal, 3, 3)
Vedere anche