假设我有4个表A,B,C,D
,每个表中都有记录和datetime
字段。
目前正在使用UNION ALL
和ORDER BY Date DESC
。
问题是如果表A
今天添加了10条记录,而表B
昨天添加了10条记录,那么前10条结果全部来自表A
。
是否可以获取按日期排序的记录,但是在所有4个表之间混合?
第一条记录将是表格A
中添加的最新记录,然后是表格B
中添加的最新记录,然后是C
,然后是D
。并像那样循环它们?
感谢
答案 0 :(得分:1)
我不确定这是你在寻找什么,但我认为你可以使用这样的东西:
SELECT
A.*,
datediff((select max(`date`) from A), `date`) as d_diff,
1 as tab
FROM A
UNION ALL
SELECT
B.*,
datediff((select max(`date`) from B), `date`) as d_diff,
2 as tab
FROM B
ORDER BY d_diff, tab, `date` desc
或者这个:
SELECT * FROM (SELECT A.* FROM A ORDER BY `date` desc limit 5) a
UNION ALL
SELECT * FROM (SELECT B.* FROM B ORDER BY `date` desc limit 5) b
ORDER BY `date` DESC
根据您的评论,这应该有效,但也可能很慢:
SELECT
A.*,
(select count(*) from A A1 where A1.dat>A.dat) as i,
1 as tab
FROM A
UNION ALL
SELECT
B.*,
(select count(*) from B B1 where B1.dat>B.dat) as i,
2 as tab
FROM B
order by i, tab
(使用变量可以更快,但我更喜欢普通的SQL)
答案 1 :(得分:1)
不幸的是,MySQL没有排名功能 - 但你可以用变量来模拟它。我建议尝试以下内容:
SELECT * FROM
(select *, @rownum := @rownum + 1 AS rank, 'A' as tbl
from (SELECT * FROM (SELECT A.* FROM A ORDER BY `date` desc) sa) a,
(SELECT @rownum := 0) r
union all
select *, @rownum2 := @rownum2 + 1 AS rank, 'B' as tbl
from (select * from (SELECT B.* FROM B ORDER BY `date` desc) sb) b,
(SELECT @rownum2 := 0) r2
union all
select *, @rownum3 := @rownum3 + 1 AS rank, 'C' as tbl
from (select * from (SELECT C.* FROM C ORDER BY `date` desc) sc) c,
(SELECT @rownum3 := 0) r3
union all
select *, @rownum4 := @rownum4 + 1 AS rank, 'D' as tbl
from (select * from (SELECT D.* FROM D ORDER BY `date` desc) sd) d,
(SELECT @rownum4 := 0) r4
) s
order by rank, tbl
或者,尝试类似:
SELECT * FROM
(SELECT S.*,
@rownum := CASE WHEN @tblvar <> S.tbl THEN 0 ELSE @rownum+1 END AS rank,
@tblvar := S.tbl AS tblvar
FROM
(SELECT * FROM
(SELECT A.*, 'A' as tbl FROM A
UNION ALL
SELECT B.*, 'B' as tbl FROM B
UNION ALL
SELECT C.*, 'C' as tbl FROM C
UNION ALL
SELECT D.*, 'D' as tbl FROM D) U
ORDER BY tbl, `date` desc) S,
(SELECT @rownum := 0) R,
(SELECT @tblvar := '!') T
) Q
order by rank, tbl