XML data as history of records

Question

I have one XML column in SQL Server where I store actual value and immediate previous value of attributes. eg. Name

<attribute name="Name">
      <actuals>
          <element isPreferred="true" name="FirstName">Name 2</element>
          <element isPreferred="false" name="LastName">N2</element>
       </actuals>
      <previous>
          <element isPreferred="true" name="FirstName">Name 1</element>
          <element isPreferred="false" name="LastName">N1</element>
      </previous>
</attribute>

How can I show the history of attribute like

------------------------------------
Attribute | New Value | OldValue
------------------------------------
First Name| Name2     | Name1
Last Name | N2        | N1
------------------------------------

Attributes may vary and can have single element like gender or multiple elements like Name or Address(addressLine1, city, state, country)

<attribute name="Gender">
      <actuals>
          <element isPreferred="true" name="Gender">Male</element>             
       </actuals>
      <previous>
          <element isPreferred="true" name="Gender">Other</element>     
      </previous>
</attribute>

score 2 · Answer 1 · answered Jan 11 '15 at 17:44

Try this.

DECLARE @xml1 XML='<attribute name="Name">
      <actuals>
          <element isPreferred="true" name="FirstName">Name 2</element>
          <element isPreferred="false" name="LastName">N2</element>
       </actuals>
      <previous>
          <element isPreferred="true" name="FirstName">Name 1</element>
          <element isPreferred="false" name="LastName">N1</element>
      </previous>
</attribute>'

SELECT Attribute=[Xml_Tab].[Cols].value('(actuals/element/@name)[1]', 'varchar(50)'),
       [New Value]=[Xml_Tab].[Cols].value('(actuals/element)[1]', 'varchar(50)'),
       OldValue=[Xml_Tab].[Cols].value('(previous/element)[1]', 'varchar(50)')
FROM   @xml1.nodes('/attribute')AS [Xml_Tab]([Cols])
UNION
SELECT [Xml_Tab].[Cols].value('(actuals/element/@name)[2]', 'varchar(50)'),
       [Xml_Tab].[Cols].value('(actuals/element)[2]', 'varchar(50)'),
       [Xml_Tab].[Cols].value('(previous/element)[2]', 'varchar(50)')
FROM   @xml1.nodes('/attribute')AS [Xml_Tab]([Cols])

Thanks, it works for Name, but Attributes may have multiple elements like Address(addressLiine1, city, state, country) — Biju Thomas, Jan 12 '15 at 06:19

Roman Pekar · Accepted Answer · 2015-01-13T12:09:18.300

You can query actual and previous nodes and then join them like this:

;with cte_act as (
    select
        t.c.value('@name', 'nvarchar(128)') as [Attribute],
        t.c.value('.', 'nvarchar(128)') as [Value]
    from @data.nodes('/attribute/actuals/element') as t(c)
), cte_prev as (
    select
        t.c.value('@name', 'nvarchar(128)') as [Attribute],
        t.c.value('.', 'nvarchar(128)') as [Value]
    from @data.nodes('/attribute/previous/element') as t(c)
)
select
    act.[Attribute],
    act.[Value] as [New Value],
    prev.[Value] as [Old Value]
from cte_act as act
    left outer join cte_prev as prev on prev.[Attribute] = act.[Attribute]

sql fiddle demo

Or you can do something crazy like reformatting your xml with xquery and then query it:

;with cte as (
    select
        t.c.query('
            for $act in actuals/element
                return <element name="{$act/@name}" newvalue="{$act/text()}" oldvalue="{($act/../../previous/element[@name=$act/@name])[1]/text()}"/>
        ') as data
    from <your table> as d
        outer apply d.data.nodes('/attribute') as t(c)
)
select
    t.c.value('@name', 'nvarchar(128)') as [Attribute],
    t.c.value('@newvalue', 'nvarchar(128)') as [New Value],
    t.c.value('@oldvalue', 'nvarchar(128)') as [Old Value]
from cte as d
    outer apply data.nodes('element') as t(c)

sql fiddle demo

Or, something like this - querying all actual elements and for each actual element try to get previous:

select
    t.c.value('@name', 'nvarchar(128)') as [Attribute],
    t.c.value('.', 'nvarchar(128)') as [Value],
    t.c.value('let $name:=@name return (../../previous/element[@name=$name])[1]', 'nvarchar(128)') as [Value]
from @temp as d
    outer apply d.data.nodes('/attribute/actuals/element') as t(c)

sql fiddle demo

Thanks Roman. Could you please suggest, from the performance stand point which one is best approach? — Biju Thomas, Jan 15 '15 at 11:12
@BijuThomas I'd suggest to use last one, but it's better to test it on your data — Roman Pekar, Jan 18 '15 at 10:53

XML data as history of records

2 Answers2