Метод STWithin
Determines whether the current SqlGeometry is completely within the specified SqlGeometry instance.
Пространство имен: Microsoft.SqlServer.Types
Сборка: Microsoft.SqlServer.Types (в Microsoft.SqlServer.Types.dll)
Синтаксис
'Декларация
<SqlMethodAttribute(IsDeterministic := True, IsPrecise := False)> _
Public Function STWithin ( _
other As SqlGeometry _
) As SqlBoolean
'Применение
Dim instance As SqlGeometry
Dim other As SqlGeometry
Dim returnValue As SqlBoolean
returnValue = instance.STWithin(other)
[SqlMethodAttribute(IsDeterministic = true, IsPrecise = false)]
public SqlBoolean STWithin(
SqlGeometry other
)
[SqlMethodAttribute(IsDeterministic = true, IsPrecise = false)]
public:
SqlBoolean STWithin(
SqlGeometry^ other
)
[<SqlMethodAttribute(IsDeterministic = true, IsPrecise = false)>]
member STWithin :
other:SqlGeometry -> SqlBoolean
public function STWithin(
other : SqlGeometry
) : SqlBoolean
Параметры
- other
Тип: Microsoft.SqlServer.Types. . :: . .SqlGeometry
The SqlGeometry instance to compare to the current SqlGeometry.
Возвращаемое значение
Тип: System.Data.SqlTypes. . :: . .SqlBoolean
Returns true if a SqlGeometry instance is completely within another SqlGeometry instance; otherwise, returns false.
Замечания
This method always returns null Nothing nullptr unit пустая ссылка (Nothing в Visual Basic) if the spatial reference IDs (SRIDs) of the SqlGeometry instances do not match.