了多个子查询的SQL查询

Question

我试图做一个SQL查询，这将有助于我acheive以下结果：

---------------------------------------------------------------------------- 
| RowNum |  email  | point_1 | point_2 | total_point | 
---------------------------------------------------------------------------- 
| 1 | abc@gmail.com |  120  |  70  |  190  | 
---------------------------------------------------------------------------- 

SQL查询语句1（以获得ROWNUM，电子邮件和_1的值） ：

$sql = " 
      select * 
      from 
      (
       select ROW_NUMBER() OVER (ORDER BY m.first_name) as **RowNum**, 
       ltrim(rtrim(m.email_addr)) AS **email**, 
       CAST(isnull(p.points_accumulated,'0') AS INT) AS **point_1** 

       FROM (select * from crm_member_list where coy_id='HSG' and mbr_id not in (select mbr_id from o2o_tmp_mbr_issues_exclude)) m 
        left join (select * from crm_member_points where coy_id='HSG') p 
        on p.mbr_id = m.mbr_id 
        where m.email_addr = 'abc@gmail.com' 

        and m.date BETWEEN '2016-08-01 00:00:00' AND '2016-08-31 23:59:00' 
       )sub where RowNum>? and RowNum<? order by RowNum"; 

SQL查询语句2（以获得point_2的值）：

$sql = " 
      select CAST(isnull(p.points_accumulated,'0') AS INT) AS **point_2** 
        FROM (select * from crm_member_list where coy_id='HSG' and mbr_id not in (select mbr_id from o2o_tmp_mbr_issues_exclude)) m 
          left join (select * from crm_member_points where coy_id='HSG') p 
          on p.mbr_id = m.mbr_id 
          where m.email_addr = 'abc@gmail.com' 
          and m.date BETWEEN '2016-09-01 00:00:00' AND '2016-09-30 23:59:00'"; 

我试图将2秒结合上面显示得到结果，但我得到的错误

“直接执行SQL，没有游标”。

组合代码：

$sql = " 
       select * 
       from 
       (
        (select ROW_NUMBER() OVER (ORDER BY m.first_name) as **RowNum**, 
        ltrim(rtrim(m.email_addr)) AS **email**, 
        CAST(isnull(p.points_accumulated,'0') AS INT) AS **point_1** 

        FROM (select * from crm_member_list where coy_id='HSG' and mbr_id not in (select mbr_id from o2o_tmp_mbr_issues_exclude)) m 
         left join (select * from crm_member_points where coy_id='HSG') p 
         on p.mbr_id = m.mbr_id 
         where m.email_addr = 'abc@gmail.com' 
         and m.date BETWEEN '2016-08-01 00:00:00' AND '2016-08-31 23:59:00'), 





        (select CAST(isnull(p.points_accumulated,'0') AS INT) AS **point_2** 
        FROM (select * from crm_member_list where coy_id='HSG' and mbr_id not in (select mbr_id from o2o_tmp_mbr_issues_exclude)) m 
          left join (select * from crm_member_points where coy_id='HSG') p 
          on p.mbr_id = m.mbr_id 
          where m.email_addr = 'abc@gmail.com' 
          and m.date BETWEEN '2016-09-01 00:00:00' AND '2016-09-30 23:59:00' 
        ) 
       )sub where RowNum>? and RowNum<? order by RowNum"; 

我应该怎样去组合这两个查询语句来生成上面显示的结果呢？如何添加point_1和point_2的列以获取total_point列？

在此先感谢

Answer 1

如果我正确理解你的问题（假设你不使用mysql，因为它不支持row_number），一种方法是使用conditional aggregation。我也觉得你实际上是在寻找sum点每月对从各月选择1：

select *, row_number() over (order by first_name) rn 
from (
    select 
     m.first_name, 
     ltrim(rtrim(m.email_addr)) AS email, 
     sum(case when m.date >= '2016-08-01' AND m.Date < '2016-09-01' 
       then CAST(isnull(p.points_accumulated,'0') AS INT) 
      end 
      ) as point_1, 
     sum(case when m.date >= '2016-09-01' AND m.Date < '2016-10-01' 
       then CAST(isnull(p.points_accumulated,'0') AS INT) 
      end 
      ) as point_2, 
     sum(CAST(isnull(p.points_accumulated,'0') AS INT)) 
    from crm_member_list m 
     left join crm_member_points p on m.coy_id = p.coy_id 
            and p.mbr_id = m.mbr_id 
    where m.coy_id = 'HSG' 
      and m.mbr_id not in (select mbr_id from o2o_tmp_mbr_issues_exclude) 
      and m.date >= '2016-08-01' AND m.date < '2016-10-01' 
    group by 1, 2 
) t 

请注意，我已经改变了你的between语句中使用<和>=。这种方法出现问题的可能性较小。