Question

我有以下表格：

CREATE table table1 (id int , cd date, ct TIME, co text)

INSERT INTO table1 
VALUES (0, '1/1/2018', '12:00:00', 'B'), 
        (1, '1/1/2018', '12:30:00', 'BC'), 
        (2, '1/12/2018', '12:00:00', 'B'), 
        (3, '1/22/2018', '12:00:00', 'BC')

当“cd”和“ct”列的值对于'B'和'BC'相同或者只有“ct”的值不同并且显示记录'B'的值时，我需要组合“co”列对于'B-BC'。对于上面的表1记录，我需要以下结果：

"id"    "cd"    "ct"    "co"
"0" "1/1/2018"  "12:00:00 PM"   "B-BC"
"2" "1/12/2018" "12:00:00 PM"   "B"
"3" "1/22/2018" "12:00:00 PM"   "BC"

在postgresql 8.4中最有效的方法是什么？

Answer 1

我创建了两个公用表表达式（CTE）来提供将要组合的B和BC记录列表。逻辑是得到B记录和BC记录。然后执行从table1到cte_A表的左连接，并将join连接到cte_B表，但cte_b.id_del为null。这将删除cte_B中找到的id。最后，做一个何时对cte_A表中找到的id使用新的co值（B-BC）的情况。请参阅此处的演示：http://sqlfiddle.com/#!15/62caa/49

with cte_a as (select a.id as id_keep 
    from table1 a 
    inner join table1 b on a.cd=b.cd
     and a.co='B' and b.co='BC') 
,cte_b  as (select b.id as id_del 
    from table1 a 
    inner join table1 b on a.cd=b.cd
    and a.co='B' and b.co='BC')
select t1.id, 
       t1.cd, 
       t1.ct, 
       case when cte_a.id_keep is null 
            then t1.co 
            else 'B-BC' end as co 
from table1 t1
left join cte_a 
on t1.id=cte_a.id_keep
left join cte_b
on t1.id = cte_b.id_del
where cte_b.id_del is null;

Answer 2

您可以使用string_agg：

SELECT cd, MIN(id) AS id, min(ct) AS ct, string_agg(co, '-' ORDER BY ct) AS co
FROM table1
GROUP BY cd;

<强> DBFiddle Demo

修改

可爱的答案！但是对于＆＃39; B＆＃39;并不总是最低价值。我需要ct，这是专门记录在＆＃39; B＆＃39;，它可能更大＆＃39;＆＃39;对于BC＆＃39;或更少

您可以使用窗口函数：

WITH cte AS ( SELECT *, ROW_NUMBER() OVER(PARTITION BY cd ORDER BY ct ASC) AS rn FROM table1 ) SELECT id, cd, ct,s.co FROM cte JOIN LATERAL(SELECT string_agg(co, '-' ORDER BY ct) AS co FROM cte c2 WHERE c2.cd=cte.cd)s ON TRUE WHERE rn=1;

<强> DBFiddle Demo2

如何在特定条件下组合两行

2 个答案: