mysql pivot table date (vertical to horizontal data)

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日久生厌 2020-12-15 01:50

I have been searching for hours without a decent answer.

I want to transform this table:


Client_id    Date
-----------  ------------  
1                


        
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  • 2020-12-15 02:35

    In order to get this result, you will want to pivot the data. MySQL does not have a pivot function but you can use an aggregate function with a CASE expression.

    If the number of dates is known, then you can hard-code the query:

    select client_id,
      max(case when rownum = 1 then date end) Date1,
      max(case when rownum = 2 then date end) Date2,
      max(case when rownum = 3 then date end) Date3
    from
    (
      select client_id,
        date,
        @row:=if(@prev=client_id, @row,0) + 1 as rownum,
        @prev:=client_id 
      from yourtable, (SELECT @row:=0, @prev:=null) r
      order by client_id, date
    ) s
    group by client_id
    order by client_id, date
    

    See SQL Fiddle with Demo

    I implemented user variables to assign a row number to each record within the client_id group.

    If you have an unknown number of dates, then you will need to use a prepared statement to create the sql dynamically:

    SET @sql = NULL;
    SELECT
      GROUP_CONCAT(DISTINCT
        CONCAT(
          'MAX(CASE WHEN rownum = ',
          rownum,
          ' THEN date END) AS Date_',
          rownum
        )
      ) INTO @sql
    from
    (
      select client_id,
        date,
        @row:=if(@prev=client_id, @row,0) + 1 as rownum,
        @prev:=client_id 
      from yourtable, (SELECT @row:=0) r
      order by client_id, date
    ) s
    order by client_id, date;
    
    
    SET @sql 
      = CONCAT('SELECT client_id, ', @sql, ' 
               from
               (
                 select client_id,
                   date,
                   @row:=if(@prev=client_id, @row,0) + 1 as rownum,
                   @prev:=client_id 
                 from yourtable, (SELECT @row:=0) r
                 order by client_id, date
               ) s
               group by client_id
               order by client_id, date');
    
    PREPARE stmt FROM @sql;
    EXECUTE stmt;
    DEALLOCATE PREPARE stmt;
    

    See SQL Fiddle with Demo.

    They both give the result:

    | CLIENT_ID |                          DATE_1 |                          DATE_2 |                     DATE_3 |
    --------------------------------------------------------------------------------------------------------------
    |         1 | February, 03 2013 00:00:00+0000 | February, 10 2013 00:00:00+0000 | May, 12 2013 00:00:00+0000 |
    |         2 | February, 03 2013 00:00:00+0000 |     July, 15 2013 00:00:00+0000 |                     (null) |
    
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