Question

我有一个名为list的表，我必须提取以下内容。字段是id和日期

如果日期＆lt; 20120401，则依靠id＆gt;＆gt;＆gt;＆gt;给结果
如果日期＆gt; 20120401，则依靠id＆gt;＆gt;＆gt;＆gt;给结果

如果有20项日期＆lt; 20120401和30项日期＆gt; 20120401的idnumber xyz

那么结果应该是......

xyz 20 30

我做得像......

select 
 (select count(id) from list where id='xyz' and date<20120401) as date1,
 (select count(id) from list where id='xyz' and date>20120401) as date2;

结果是20 30

但是如何打印idnumber？

Answer 1

SELECT
id,
SUM(CASE WHEN date < 20120401 THEN 1 ELSE 0 END) AS date1,
SUM(CASE WHEN date > 20120401 THEN 1 ELSE 0 END) AS date2,
FROM list
WHERE id = 'xyz'
GROUP BY id

更新：

SELECT
list.id,
idmaster.idlocation,
SUM(CASE WHEN list.date < 20120401 THEN 1 ELSE 0 END) AS date1,
SUM(CASE WHEN list.date > 20120401 THEN 1 ELSE 0 END) AS date2,
FROM list
INNER JOIN idmaster ON list.id = idmaster.idnumber
WHERE list.id = 'xyz'
GROUP BY id

Answer 2

尝试

select id, count(id) from list where id='xyz' and date < 20120401
union
select id, count(id) from list where id='xyz' and date > 20120401

Answer 3

您正在寻找GROUP BY条款，可能是这样的：

SELECT
    l2.id,
    (SELECT COUNT(id)
    FROM list l1
    WHERE l1.id = l2.id AND date < 20120401) AS date1,
    (SELECT COUNT(id)
    FROM list l1
    WHERE l1.id = l2.id AND date > 20120401) AS date2
FROM
    list l2
GROUP BY
    l2.id

有更多具有成本效益的方法可以获得理想的结果，但这是最容易理解的。

Answer 4

试试这个：

 SELECT
    id,
    SUM(CASE WHEN date < 20120401 
    THEN 1 ELSE 0 END) AS date1,
    SUM(CASE WHEN date > 20120401 
    THEN 1 ELSE 0 END) AS date2,
    FROM list
    WHERE id = 'xyz'
    GROUP BY id

多个选择在一个表上

4 个答案: