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  4. SQL server 12: Displaying column cell values in single cell

SQL server 12: Displaying column cell values in single cell

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  • S Swap9

    Column1 Column2 -------- ---------- Vaugh William Vaugh Smith Woods Jane Expected Output : Vaugh, William-Smith/Woods,Jane Condition : There could be n number of rows in the sample table, there could be n number of woods too. display all the common surname having different names meaning repeated Last names will be once then all the first names associated with it. truncated at 30 characters max. How can I get this output in SQL query ?

    Richard DeemingR Offline
    Richard DeemingR Offline
    Richard Deeming
    wrote on last edited by
    #8

    This sort of manipulation is best done in the UI, not the database. However, using the techniques from this blog post[^] leads to the following (rather ugly) solution:

    SELECT
    STUFF(
    (SELECT '/' + Surname + ',' + STUFF(
    (SELECT '-' + Forename
    FROM YourTable As T2
    WHERE T2.Surname = T1.Surname
    ORDER BY T2.Forename
    FOR XML PATH(''), TYPE
    ).value('.', 'varchar(max)')
    , 1, 1, '')
    FROM YourTable As T1
    GROUP BY Surname
    ORDER BY Surname
    FOR XML PATH(''), TYPE
    ).value('.', 'varchar(max)')
    , 1, 1, '') As CombinedNames
    ;


    "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

    "These people looked deep within my soul and assigned me a number based on the order in which I joined" - Homer

    S 1 Reply Last reply
    0
    • S Swap9

      I want to display the unique last names followed by their respective first names separated by a '-' and, then '/' , then again the next last name(If it is not distinct , then add their first names separated by '-')followed by its respective firstname and so on..

      L Offline
      L Offline
      Lost User
      wrote on last edited by
      #9

      How do you know which first name belongs to which last name?

      Bastard Programmer from Hell :suss: If you can't read my code, try converting it here[^]

      1 Reply Last reply
      0
      • Richard DeemingR Richard Deeming

        This sort of manipulation is best done in the UI, not the database. However, using the techniques from this blog post[^] leads to the following (rather ugly) solution:

        SELECT
        STUFF(
        (SELECT '/' + Surname + ',' + STUFF(
        (SELECT '-' + Forename
        FROM YourTable As T2
        WHERE T2.Surname = T1.Surname
        ORDER BY T2.Forename
        FOR XML PATH(''), TYPE
        ).value('.', 'varchar(max)')
        , 1, 1, '')
        FROM YourTable As T1
        GROUP BY Surname
        ORDER BY Surname
        FOR XML PATH(''), TYPE
        ).value('.', 'varchar(max)')
        , 1, 1, '') As CombinedNames
        ;


        "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

        S Offline
        S Offline
        Swap9
        wrote on last edited by
        #10

        I am getting the combined names for the whole table. How to restrict it for a single id? Can you please tell where should I put the condition in the code you have shown above ?

        Richard DeemingR 1 Reply Last reply
        0
        • S Swap9

          I am getting the combined names for the whole table. How to restrict it for a single id? Can you please tell where should I put the condition in the code you have shown above ?

          Richard DeemingR Offline
          Richard DeemingR Offline
          Richard Deeming
          wrote on last edited by
          #11

          What do you mean by "a single id"? There's no mention of any column called "id" in your question, or any of the other messages in this thread.


          "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

          "These people looked deep within my soul and assigned me a number based on the order in which I joined" - Homer

          S 1 Reply Last reply
          0
          • Richard DeemingR Richard Deeming

            What do you mean by "a single id"? There's no mention of any column called "id" in your question, or any of the other messages in this thread.


            "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

            S Offline
            S Offline
            Swap9
            wrote on last edited by
            #12

            ID Column1 Column2 -- -------- ---------- 7 Vaugh William 7 Vaugh Smith 6 Woods Jane 6 Woods Joseph 6 Wright Adam 6 Wright John Suppose I have to find the combined name only for id=6 , then where should I put the condition in the code? The expected output : Woods,Jane-Joseph/Wright,Adam-John

            Richard DeemingR 1 Reply Last reply
            0
            • S Swap9

              ID Column1 Column2 -- -------- ---------- 7 Vaugh William 7 Vaugh Smith 6 Woods Jane 6 Woods Joseph 6 Wright Adam 6 Wright John Suppose I have to find the combined name only for id=6 , then where should I put the condition in the code? The expected output : Woods,Jane-Joseph/Wright,Adam-John

              Richard DeemingR Offline
              Richard DeemingR Offline
              Richard Deeming
              wrote on last edited by
              #13

              So the ID is the same for each surname? In that case, you just need to add a WHERE clause between the FROM YourTable As T1 and GROUP BY Surname lines:

              SELECT
              STUFF(
              (SELECT '/' + Surname + ',' + STUFF(
              (SELECT '-' + Forename
              FROM YourTable As T2
              WHERE T2.Surname = T1.Surname
              ORDER BY T2.Forename
              FOR XML PATH(''), TYPE
              ).value('.', 'varchar(max)')
              , 1, 1, '')
              FROM YourTable As T1

                  -- Filter here:
                  WHERE T1.ID = @TheIDToFind
                  
                  GROUP BY Surname
                  ORDER BY Surname
                  FOR XML PATH(''), TYPE
              ).value('.', 'varchar(max)')
              , 1, 1, '') As CombinedNames
              

              ;


              "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

              "These people looked deep within my soul and assigned me a number based on the order in which I joined" - Homer

              S 1 Reply Last reply
              0
              • Richard DeemingR Richard Deeming

                So the ID is the same for each surname? In that case, you just need to add a WHERE clause between the FROM YourTable As T1 and GROUP BY Surname lines:

                SELECT
                STUFF(
                (SELECT '/' + Surname + ',' + STUFF(
                (SELECT '-' + Forename
                FROM YourTable As T2
                WHERE T2.Surname = T1.Surname
                ORDER BY T2.Forename
                FOR XML PATH(''), TYPE
                ).value('.', 'varchar(max)')
                , 1, 1, '')
                FROM YourTable As T1

                    -- Filter here:
                    WHERE T1.ID = @TheIDToFind
                    
                    GROUP BY Surname
                    ORDER BY Surname
                    FOR XML PATH(''), TYPE
                ).value('.', 'varchar(max)')
                , 1, 1, '') As CombinedNames
                

                ;


                "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

                S Offline
                S Offline
                Swap9
                wrote on last edited by
                #14

                ID Column1 Column2 -- -------- ---------- 7 Vaugh William 7 Vaugh Smith 6 Woods Jane 6 Woods Joseph 6 Wright Adam 6 Wright John Thanks, I exactly put the condition at the place you suggeted, but still the output I get is as below WRONG OP : Vaugh,William-Smith/Woods,Jane-Joseph/Wright, Adam-John Expected OP : Woods,Jane-Joseph/Wright, Adam-John

                Richard DeemingR 2 Replies Last reply
                0
                • S Swap9

                  ID Column1 Column2 -- -------- ---------- 7 Vaugh William 7 Vaugh Smith 6 Woods Jane 6 Woods Joseph 6 Wright Adam 6 Wright John Thanks, I exactly put the condition at the place you suggeted, but still the output I get is as below WRONG OP : Vaugh,William-Smith/Woods,Jane-Joseph/Wright, Adam-John Expected OP : Woods,Jane-Joseph/Wright, Adam-John

                  Richard DeemingR Offline
                  Richard DeemingR Offline
                  Richard Deeming
                  wrote on last edited by
                  #15

                  I've just tried the code here on your sample data, and I get the expected output.

                  DECLARE @T TABLE
                  (
                  ID int NOT NULL,
                  Surname varchar(10) NOT NULL,
                  Forename varchar(10) NOT NULL
                  );

                  INSERT INTO @T (ID, Surname, Forename)
                  VALUES
                  (7, 'Vaugh', 'William'),
                  (7, 'Vaugh', 'Smith'),
                  (6, 'Woods', 'Jane'),
                  (6, 'Woods', 'Joseph'),
                  (6, 'Wright', 'Adam'),
                  (6, 'Wright', 'John')
                  ;

                  SELECT
                  STUFF(
                  (SELECT '/' + Surname + ',' + STUFF(
                  (SELECT '-' + Forename
                  FROM @T As T2
                  WHERE T2.Surname = T1.Surname
                  ORDER BY T2.Forename
                  FOR XML PATH(''), TYPE
                  ).value('.', 'varchar(max)')
                  , 1, 1, '')
                  FROM @T As T1
                  WHERE T1.ID = 6
                  GROUP BY Surname
                  ORDER BY Surname
                  FOR XML PATH(''), TYPE
                  ).value('.', 'varchar(max)')
                  , 1, 1, '') As CombinedNames
                  ;

                  -- Output:
                  -- Woods,Jane-Joseph/Wright,Adam-John


                  "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

                  "These people looked deep within my soul and assigned me a number based on the order in which I joined" - Homer

                  1 Reply Last reply
                  0
                  • S Swap9

                    ID Column1 Column2 -- -------- ---------- 7 Vaugh William 7 Vaugh Smith 6 Woods Jane 6 Woods Joseph 6 Wright Adam 6 Wright John Thanks, I exactly put the condition at the place you suggeted, but still the output I get is as below WRONG OP : Vaugh,William-Smith/Woods,Jane-Joseph/Wright, Adam-John Expected OP : Woods,Jane-Joseph/Wright, Adam-John

                    Richard DeemingR Offline
                    Richard DeemingR Offline
                    Richard Deeming
                    wrote on last edited by
                    #16

                    Here's a SQL Fiddle with the same query, which also generates the correct output: http://sqlfiddle.com/#!3/300af/1[^]


                    "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

                    "These people looked deep within my soul and assigned me a number based on the order in which I joined" - Homer

                    S 1 Reply Last reply
                    0
                    • Richard DeemingR Richard Deeming

                      Here's a SQL Fiddle with the same query, which also generates the correct output: http://sqlfiddle.com/#!3/300af/1[^]


                      "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

                      S Offline
                      S Offline
                      Swap9
                      wrote on last edited by
                      #17

                      I had to add few extra condition suitable to my table, but it DID WORK !! :) I have learnt about STUFF and FOR XML PATH. Thanks a ton for all your time Richard.

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