【发布时间】:2013-01-29 09:56:20
【问题描述】:
请有人帮我解决我已经为此花了两天时间的 sql 查询....
我在下面给出了一个 MYSQL 查询
(SELECT
c.cl_list as cl_list,
c.name as name,
pc.value as value,
count( pc.value) as total
FROM
projs p
LEFT JOIN classify_proj_new pc
ON p.proj_id = pc.proj_id_fk
LEFT JOIN classify_list c
ON c.cl_list = pc.class_id_fk
WHERE
MATCH ( p.title ) AGAINST ( 'jerm' IN BOOLEAN MODE )
GROUP BY
c.cl_list,
pc.value)
UNION ALL
(SELECT
c.cl_list as cl_list,
c.name as name,
pc.value as value,
count( pc.value) as total
FROM
jerm p
LEFT JOIN classify_jerm_new pc
ON p.jerm_id = pc.jerm_id_fk
LEFT JOIN classify_list c
ON c.cl_list = pc.class_id_fk
WHERE
MATCH ( p.jermname ) AGAINST ( 'jerm' IN BOOLEAN MODE )
GROUP BY
c.cl_list,
pc.value)
结果如下:
cl_list name value total
------------------------------------------------------------------------------------
1 department jewller 2
3 price 50 2
6 color blue 1
6 color Red 2
1 department jewller 1
6 color Red 1
但我试图得到一个结果,它可以添加重复值的总数并避免重复值....类似这样的事情(下):
cl_list name value total
------------------------------------------------------------------------------------
1 department jewller 3
3 price 50 2
6 color blue 1
6 color Red 3
请有人帮助我,我对我的输出感到非常难过...
非常感谢您提前...
【问题讨论】:
-
groupname 和 value 的结果,而不是 cl_list 和 value, -
@AkamOmer 我尝试过这种方式,但仍然得到相同的结果。what todo.thank
-
@SuhelMeman 伙计,我以另一种方式得到相同的结果..任何其他服从...thanx
-
检查下面的查询,我在答案部分发布了