GROUP BY部分

Question

我有桌子：

ID_table, User, Multi_ID, Action, Date, ... 
100     , JN    , 0     , test1 , 1.1.2017
101     , DK    , 51    , test2 , 3.1.2017
102     , JN    , 51    , test2 , 3.1.2017
103     , LP    , 0     , test3 , 3.1.2017
104     , DK    , 52    , test4 , 5.1.2017
105     , LP    , 52    , test4 , 5.1.2017
106     , KH    , 52    , test4 , 5.1.2017

我想制作GROUP BY Multi_ID，但仅适用于Multi_ID＆gt; 0。 如果Multi_ID = 0 - 没有GROUP BY ...：

Date      Users      Action
1.1.2017  (JN)       test1
3.1.2017  (DK,JN)    test2
3.1.2017  (LP)       test3
5.1.2017  (DK,LP,KH) test4

我的代码没有解决方案Multi_ID = 0

$sql="SELECT * , GROUP_CONCAT(DISTINCT User separator ',' ) AS Us    
      FROM table
  GROUP BY Multi_ID
  ORDER BY Date ASC";
$result = MySQL_Query($sql);

由于

Answer 1

您需要两个查询，其结果与UNION ALL：

粘合在一起

SELECT date, user AS users, action
FROM mytable
WHERE multi_id = 0
UNION ALL
SELECT 
  MAX(date) AS date, 
  GROUP_CONCAT(DISTINCT user SEPARATOR ',' ) AS users,
  MAX(action) AS action
FROM mytable
WHERE multi_id > 0
GROUP BY multi_id
ORDER BY date;

另一种选择是在UUID的情况下按Multi_ID = 0进行分组，以便将它们分开。这可能更优雅，更不容易理解; - ）

SELECT 
  MAX(date) as date, 
  GROUP_CONCAT(DISTINCT user SEPARATOR ',' ) AS users,
  MAX(action) as action
FROM mytable
GROUP BY CASE WHEN multi_id = 0 THEN uuid() ELSE multi_id END
ORDER BY date;