我有以下[表格a]
id res1 res2
1 a f
1 b f
1 b f
1 c f
2 e g
2 e g
2 e g
2 f g
我在执行group_concat后得到以下内容
select
id,
group_concat(case when cnt = 1 then res1 else concat(cnt, ' ', res1) end) as r1,
group_concat(case when cnt = 1 then res2 else concat(cnt, ' ', res2) end) as r2
from
(
select id, res1,res2, count(*) as cnt
from [table a]
group by id, res1,res2
) t
group by id;
id r1 r2
1 a,2 b,c f,2 f,f
2 3 e,f 3 g,g
res1列正常,但res2列正在复制res1列。 基本上我想打印一个角色出现在角色之前的次数值。 我希望采用以下格式..
id r1 r2
1 a,2 b,c 4 f
2 3 e,f 4 g
我怎样才能实现它?
答案 0 :(得分:1)
我接近这个的方法是使用两个单独的子查询对res1
和res2
列进行两次汇总/聚合。第一个聚合超过id
和 res1
(或res2
),并获取每个字母或单词的计数。然后,再次聚合,这次仅在id
上,以获得每个id
的逗号分隔字符串。最后,将这些子查询连接在一起以获得最终结果。
SELECT
t1.id, t1.r1, t2.r2
FROM
(
SELECT t.id, GROUP_CONCAT(res1agg ORDER BY res1) AS r1
FROM
(
SELECT
id,
res1,
CASE WHEN COUNT(*) = 1 THEN res1
ELSE CONCAT(CAST(COUNT(*) AS CHAR(50)), res1) END AS res1agg
FROM yourTable
GROUP BY id, res1
) t
GROUP BY t.id
) t1
INNER JOIN
(
SELECT t.id, GROUP_CONCAT(res2agg ORDER BY res2) AS r2
FROM
(
SELECT
id,
res2,
CASE WHEN COUNT(*) = 1 THEN res2
ELSE CONCAT(CAST(COUNT(*) AS CHAR(50)), res2) END AS res2agg
FROM yourTable
GROUP BY id, res2
) t
GROUP BY t.id
) t2
ON t1.id = t2.id;
<强>输出:强>
在这里演示:
答案 1 :(得分:0)
在查询中添加了2个条件,而不使用更多内部查询。
试试这个: -
input:-
CREATE TABLE Table1 (id INT, res1 varchar(20), res2 varchar(20));
insert into Table1 values(1, 'a', 'f');
insert into Table1 values(1, 'b', 'f');
insert into Table1 values(1, 'b', 'f');
insert into Table1 values(1, 'c', 'f');
insert into Table1 values(2, 'e', 'g');
insert into Table1 values(2, 'e', 'g');
insert into Table1 values(2, 'e', 'g');
insert into Table1 values(2, 'f', 'g');
Query:-
Select t.id,group_concat(case when cnt = 1 then res1 else concat(cnt, ' ', res1) end) as r1,
case when id=1 then trim(concat(sum(case when id = 1 then cnt end),' ',res2))
else trim(concat(sum(case when id = 2 then cnt end),' ',res2)) end as r2
from
(
select id, res1,res2,count(*) as cnt
from table1 a
group by id, res1,res2
) t
group by t.id
My Output:-
id r1 r2
1 a,c,2 b 4 f
2 f,3 e 4 g
如果您有任何问题,请告诉我