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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

    What extra information will make it clear ?

    Z Offline
    Z Offline
    ZurdoDev
    wrote on last edited by
    #4

    It looks like you want a single record as your output? What are the conditions?

    There are only 10 types of people in the world, those who understand binary and those who don't.

    S 1 Reply Last reply
    0
    • Z ZurdoDev

      It looks like you want a single record as your output? What are the conditions?

      There are only 10 types of people in the world, those who understand binary and those who don't.

      S Offline
      S Offline
      Swap9
      wrote on last edited by
      #5

      yes there is just single cell output. but conditions are that cell can not have more than 30 characters. Even there are 10 or 15 rows in a table, the SQL query should give single cell output.

      Z 1 Reply Last reply
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      • S Swap9

        yes there is just single cell output. but conditions are that cell can not have more than 30 characters. Even there are 10 or 15 rows in a table, the SQL query should give single cell output.

        Z Offline
        Z Offline
        ZurdoDev
        wrote on last edited by
        #6

        That doesn't make any sense. You want to concatenate everything into a single value? If so you can do similar to:

        DECLARE @temp NVARCHAR(MAX)
        SELECT @temp = COALESCE(@temp + ', ', '') + field1
        FROM table

        There are only 10 types of people in the world, those who understand binary and those who don't.

        S 1 Reply Last reply
        0
        • Z ZurdoDev

          That doesn't make any sense. You want to concatenate everything into a single value? If so you can do similar to:

          DECLARE @temp NVARCHAR(MAX)
          SELECT @temp = COALESCE(@temp + ', ', '') + field1
          FROM table

          There are only 10 types of people in the world, those who understand binary and those who don't.

          S Offline
          S Offline
          Swap9
          wrote on last edited by
          #7

          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 1 Reply Last reply
          0
          • 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
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            • 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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