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Home/ Questions/Q 7984979
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T11:26:31+00:00 2026-06-04T11:26:31+00:00

Please check below query. declare @xmlRoot as xml set @xmlRoot= ‘<Root> <table1 col1=2012-03-02T16:42:55.777> <table2Array>

  • 0

Please check below query.

        declare @xmlRoot as xml
        set @xmlRoot= '<Root>
        <table1 col1="2012-03-02T16:42:55.777">
            <table2Array>
              <Table2 col2="abc">
              </Table2>  
              <Table2 col2="def"> 
              </Table2> 
            </table2Array>
          </table1>
         <table1 col1="2012-03-02T17:42:55.777">
            <table2Array>
              <Table2 col2="abc1">
              </Table2>  
              <Table2 col2="def1"> 
              </Table2> 
            </table2Array>
          </table1>
        </Root>'

        declare @a as varchar(1) 
           set @a= '1' 

         SELECT
        col1 =  item.value('./@col2', 'varchar(10)') 
        FROM @xmlRoot.nodes('Root/table1[1]/table2Array/Table2'  ) AS T(item);

–The above query return expected output

SELECT
col1 =  item.value('./@col2', 'varchar(10)') 
FROM @xmlRoot.nodes('Root/table1[*[local-name()=sql:variable("@a")]]/table2Array/Table2'  ) 
  AS T(item);

–The above query doesn’t return expected output

what am I doing wrong here?

Since I dont have a key value in parent node to identify child node. I have to parse through index.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-04T11:26:32+00:00Added an answer on June 4, 2026 at 11:26 am

    This worked for me:

    DECLARE @a INT; -- data type is probably important!
    SET @a = 1;
    
    SELECT col1 =  item.value('./@col2', 'varchar(10)') 
    FROM @xmlRoot.nodes('Root/table1[sql:variable("@a")]/table2Array/Table2') AS T(item);
    
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