Ervan uitgaande dat uw voorbeeld-XML er ongeveer zo uitziet:
<MissingDS>
<MissingTable>
<MissingColumn>abc</MissingColumn>
<TableName>tblMyTable</TableName>
<PhysicalColName>table_abc</PhysicalColName>
<Grantor_Grantee>nobody</Grantor_Grantee>
</MissingTable>
<MissingTable>
<MissingColumn>xyu</MissingColumn>
<TableName>tblMyTable2</TableName>
<PhysicalColName>table_xyz</PhysicalColName>
<Grantor_Grantee>nobody2</Grantor_Grantee>
</MissingTable>
</MissingDS>
Dan zou je dit als volgt kunnen ontleden met de nieuwe SQL Server 2005 XQuery-ondersteuning:
DECLARE @MissingXML XML
SET @MissingXML = CAST(@MissingRecordsXML AS XML)
SELECT
Missing.Rec.value('(MissingColumn)[1]', 'varchar(1000)') AS 'MissingColumn',
Missing.Rec.value('(TableName)[1]', 'varchar(100)') AS 'TableName',
Missing.Rec.value('(PhysicalColName)[1]', 'varchar(100)') AS 'Physical',
Missing.Rec.value('(Grantor_Grantee)[1]', 'varchar(100)') AS 'Grantor_Grantee'
FROM
@MissingXML.nodes('/MissingDS/MissingTable') AS Missing(Rec)
Natuurlijk, als je het kunt SELECTEREN, kun je diezelfde gegevensrijen ook vrij gemakkelijk in een tabel INVOEGEN:
INSERT INTO
dbo.MissingDSTable(MissingColumn, TableName, PhysicalColName, Grantor_Grantee)
SELECT
Missing.Rec.value('(MissingColumn)[1]', 'varchar(1000)') AS 'MissingColumn',
Missing.Rec.value('(TableName)[1]', 'varchar(100)') AS 'TableName',
Missing.Rec.value('(PhysicalColName)[1]', 'varchar(100)') AS 'Physical',
Missing.Rec.value('(Grantor_Grantee)[1]', 'varchar(100)') AS 'Grantor_Grantee'
FROM
@MissingXML.nodes('/MissingDS/MissingTable') AS Missing(Rec)
Ik hoop dat dit een beetje helpt
Marc