票证条目：

id | ticket_status_id
1  | 2
2  | 3
3  | 4
4  | 3

操作日志：

id | ticket_entry_id | action_type | createtime
 1 |     1           |   2         | 2022-10-13 15:25:25.202417+00
 2 |     1           |   3         | 2022-10-11 11:25:25.202417+00
 3 |     2           |   3         | 2022-10-12 12:25:25.202417+00
 4 |     3           |   1         | 2022-10-14 15:25:25.202417+00
 5 |     4           |   3         | 2022-10-15 15:25:25.202417+00

我希望按其action_type为3的operate_log的create_time值对（3，4）ticket_entry中的所有ticket_status_id进行排序，而ticket_status_id不在（3，4）中或action_type不为3的operate_log的那些ticket_entry将保留在结果的最后
预期结果应该是

id | action_type
 4 |   3         
 2 |   3         
 1 |   2         
 3 |   4

我试过这个

SELECT DISTINCT
    ( e.* ), o.create_time
FROM
    ticket_entries e
    LEFT JOIN operate_logs o ON e.ID = o.ticket_entry_id 
    AND o.action_type = 3 
ORDER BY
CASE
        
        WHEN e.ticket_status_id IN ( 3, 4 ) THEN
        1 
    END,
    o.create_time ASC NULLS LAST

结果表明

ERROR:  for SELECT DISTINCT, ORDER BY expressions must appear in select list
LINE 10: CASE

但这句话很管用

SELECT DISTINCT
    (e.*), o.create_time
FROM
    ticket_entries e
    LEFT JOIN operate_logs o ON e.ID = o.ticket_entry_id 
    AND o.action_type = 3 
ORDER BY

    o.create_time ASC NULLS LAST

ticket_entries和operate_logs之间关系是一对多
如果我想让sql正常工作，我还应该选择什么