Question

我正在创建2个查询：

1

select count(c.id) as cid1, p.name as pname1, u.name as uname1 
from crm_lead c, account_period p, res_users u 
where c.create_date between p.date_start and p.date_stop and 
      (c.user_id = u.id or c.sales_vertical=u.id) and 
      u.id = 1 
group by p.name, u.name

2

select count(c.id) as cid2, p.name as pname2, u.name as uname2 
from crm_lead c, account_period p, res_users u 
where c.create_date between p.date_start and p.date_stop and
      (c.user_id = u.id or c.sales_vertical=u.id) and 
      stage_id =12 and 
      u.id = 1 
where a.pname = b.pname 
group by p.name, u.name

我得到以下2个查询的结果

q1的结果：

 cid1         pname1      uname1                                       
 11           07/2011     admin                                        
 5            08/2011     admin                                        
 9            09/2011     admin

q2的结果：

 cid2        pname2    uname2                                          
 9           07/2011   admin                                           
 3           09/2011   admin

将两个查询组合后我得到以下输出：

 cid1  cid2  pname1    pname2     uname1     uname2                    
 11     9     07/2011   07/2011   admin      admin                     
 9      3     09/2011   09/2011   admin      admin

但我希望结果为

 cid1   cid2  pname1    pname2    uname1     uname2                    
 11     9     07/2011   07/2011   admin      admin                     
 5      0     08/2011   08/2011   admin      admin                     
 9      3     09/2011   09/2011   admin      admin

怎么做？

我的综合查询如下：

select a.cid1, b.cid2, a.pname1, b.pname2, a.uname1, b.uname2 
from (select count(c.id) as cid1, p.name as pname1, u.name1 as uname 
      from crm_lead c, account_period p, res_users u 
      where c.create_date between p.date_start and p.date_stop and 
            (c.user_id = u.id or c.sales_vertical=u.id) and 
            u.id = 1 
      group by p.name, u.name) as a,
     (select count(c.id) as cid2, p.name as pname2, u.name as uname2 
      from crm_lead c, account_period p, res_users u 
      where c.create_date between p.date_start and p.date_stop and
            (c.user_id = u.id or c.sales_vertical=u.id) and 
            stage_id =12 and 
            u.id = 1 
      group by p.name, u.name)as b 
where a.pname1 = b.pname2

Answer 1

尝试使用LEFT JOIN：

    select a.cid1, b.cid2, a.pname1, b.pname2, a.uname1, b.uname2 from

    (select count(c.id) as cid1, p.name as pname1, u.name1 as uname from crm_lead c, account_period p, res_users u where c.create_date between p.date_start and p.date_stop and (c.user_id = u.id or c.sales_vertical=u.id) and u.id = 108 group by p.name, u.name) as a

left join

    (select count(c.id) as cid2, p.name as pname2, u.name as uname2 from crm_lead c, account_period p, res_users u where c.create_date between p.date_start and p.date_stop and(c.user_id = u.id or c.sales_vertical=u.id) and stage_id =12 and u.id = 108 group by p.name, u.name) as b on a.pname1 = b.pname2

Answer 2

要获取您请求的结果，查询可能如下所示：

SELECT a.cid1
      ,COALESCE(b.cid2, 0) AS cid2
      ,a.pname1
      ,COALESCE(b.pname2, a.pname2) AS pname2
      ,a.uname1
      ,COALESCE(b.uname2, a.uname2) AS uname2
FROM  (<query1>)  a
LEFT   JOIN (<query2>) b ON a.pname1 = b.pname2

然而，我怀疑你确实想要那个确切的结果。

Answer 3

使用ANSI-92显式连接语法，而不是旧的隐式连接语法：

select a.cid1, 
       coalesce(b.cid2,0) cid2, 
       a.pname1, 
       coalesce(b.pname2, a.pname1) pname2, 
       a.uname1, 
       coalesce(b.uname2, a.uname1) uname2
from (select count(c.id) as cid1, p.name as pname1, u.name1 as uname 
      from crm_lead c
      join account_period p on (c.create_date between p.date_start and p.date_stop)
      join res_users u on (c.user_id = u.id or c.sales_vertical=u.id) and u.id=108 
      group by p.name, u.name) as a
left join
     (select count(c.id) as cid2, p.name as pname2, u.name as uname2 
      from crm_lead c
      join account_period p on c.create_date between p.date_start and p.date_stop
      join res_users u 
        on (c.user_id = u.id or c.sales_vertical=u.id) and stage_id=12 and u.id=108 
      group by p.name, u.name) as b 
  on a.pname1 = b.pname2

如何在下面的示例中使用外连接？

3 个答案: