DataType 构造函数 (XmlSchemaCollection)
Initializes a new instance of the DataType class based on a specified XML schema collection.
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 Sub New ( _
xmlSchemaCollection As XmlSchemaCollection _
)
用法
Dim xmlSchemaCollection As XmlSchemaCollection
Dim instance As New DataType(xmlSchemaCollection)
public DataType(
XmlSchemaCollection xmlSchemaCollection
)
public:
DataType(
XmlSchemaCollection^ xmlSchemaCollection
)
new :
xmlSchemaCollection:XmlSchemaCollection -> DataType
public function DataType(
xmlSchemaCollection : XmlSchemaCollection
)
参数
- xmlSchemaCollection
类型:Microsoft.SqlServer.Management.Smo. . :: . .XmlSchemaCollection
A XmlSchemaCollection object variable that specifies an XML schema collection.
示例
Visual Basic
'Connect to the local, default instance of SQL Server.
Dim srv As Server
srv = New Server()
'Reference the AdventureWorks2008R2 database.
Dim db As Database
db = srv.Databases("AdventureWorks2008R2")
'Define an XmlSchemaCollection object by supplying the parent database and name arguments in the constructor.
Dim xsc As XmlSchemaCollection
xsc = New XmlSchemaCollection(db, "MySampleCollection")
xsc.Text = "<xsd:schema xmlns:xsd=\"http://www.w3.org/2001/XMLSchema\" targetNamespace=\"NS2\"><xsd:element name=\"elem1\" type=\"xsd:integer\"/></xsd:schema>";
'Create the XML schema collection on the instance of SQL Server.
xsc.Create()
Dim dt As DataType
dt = New DataType(xsc)
PowerShell
$srv = new-Object Microsoft.SqlServer.Management.Smo.Server("(local)")
$db = New-Object Microsoft.SqlServer.Management.Smo.Database
$db = $srv.Databases.Item("AdventureWorks2008R2")
$xsc = new-object Microsoft.SqlServer.Management.Smo.XmlSchemaCollection($db, "MySampleCollectfghion2")
$xsc.Text = "<xsd:schema xmlns:xsd=\`"http://www.w3.org/2001/XMLSchema\`" targetNamespace=\`"NS2\`"><xsd:element name=\`"elem1\`" type=\`"xsd:integer\`"/></xsd:schema>"
$xsc.Create()
$dt = new-object Microsoft.SqlServer.Management.Smo.DataType($xsc)