Добавяне/промяна на XML колона с атрибути от друга стойност на колона, връщане на резултатите като колона с други полета

declare @TableX table (ID int identity, XmlField xml)

insert into @TableX values
(
'<Node Attrib="9">Name1</Node>
 <Node Attrib="100">Name2</Node>
 <Node Attrib="101">Name2</Node>'
)

insert into @TableX values
(
'<Node Attrib="9">Name1</Node>
 <Node Attrib="101">Name2</Node>'
)

insert into @TableX values
(
'<Node Attrib="1">Name1</Node>'
)

declare @TableY table (IntField int, OtherField varchar(15))

insert into @TableY values 
(9, 'SomeOtherValue1'),
(100, 'SomeOtherValue2'),
(101, 'SomeOtherValue3')

;with C as
(
  select X.ID,
         Y.IntField as Attrib,
         Y.OtherField as OtherField
  from @TableX as X
    cross apply X.XmlField.nodes('/Node') as T(N)
    inner join @TableY as Y
      on T.N.value('@Attrib', 'int') = Y.IntField
)
select (select C2.Attrib as '@Attrib',
               C2.OtherField as '@OtherField'
        from C as C2
        where C1.ID = C2.ID
        for xml path('Node'), type) as XMLField       
from C as C1
group by ID

Резултат:

XMLField
<Node Attrib="9" OtherField="SomeOtherValue1" /><Node Attrib="100" OtherField="SomeOtherValue2" /><Node Attrib="101" OtherField="SomeOtherValue3" />
<Node Attrib="9" OtherField="SomeOtherValue1" /><Node Attrib="101" OtherField="SomeOtherValue3" />