如何在sql server 2008中选择这个xml

Question

I need to select all guid node in the xml. but the code below only select the first one of them. how to do it?

DECLARE @doc XML

SET @doc = 
    '<ArrayOfGuid xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
      <guid>96eecbe2-d645-465d-8232-c7f21e3c6bf8</guid>
      <guid>38985702-0c0b-4e9c-bccb-af84ba4dd7ff</guid>
      <guid>67852205-092e-4db8-b31e-6f5d457db294</guid>
      <guid>92cf9106-445f-4b01-8259-613596b8a2a7</guid>
    </ArrayOfGuid>'

DECLARE @docHandle INT

EXEC sp_xml_preparedocument @docHandle OUTPUT,
     @doc

SELECT [guid]
FROM   OPENXML(@docHandle, '/ArrayOfGuid', 2)
       WITH
       ([guid] UNIQUEIDENTIFIER)

the result is just one row: 96EECBE2-D645-465D-8232-C7F21E3C6BF8

I need all 4 rows.

Answer 1

There are probably better ways to do this, but here's one of them:

SELECT text as guid
FROM   OPENXML(@docHandle, '//ArrayOfGuid/guid/text()', 2)

Answer 2

You can also use this approach using the native XQuery capabilities in SQL Server 2005 and newer:

SELECT
    Guids.value('(.)[1]', 'uniqueidentifier')
FROM 
    @doc.nodes('/ArrayOfGuid/guid') AS XTbl(Guids)

I always prefer this method over the older OPENQUERY approach