Question

我有什么：

2桌

- table_benefactor ( id and benefactor names)
- table_realstate  (house adress, rent, paidrent flag)

If paidrent = 0 (unpaid)
If paidrent = 1 (paid)

我需要什么：

1个嵌套查询，返回如下结果

name | totalrent | paid | unpaid
_________________________________
Jhon | 1,000.00  |100.00| 900.00
Doe  | 2,500.00  |500.00| 2,000.00
Chris| 800.00    |0.00  | 800.00

我尝试了什么：

我尝试了很多方法，但是他们没有给我我需要的结果，他们接近但不是我需要的，这是我的疑问 - ＆gt;

如果我这样做，它将完全返回我想要的但不是我需要它的方式

(SELECT table_realstate.benefactor, name, SUM(rent) totalrent 
FROM table_realstate, table_benefactor 
WHERE flag = 0 
AND table_realstate.benefactor = table_benefactor.id
GROUP BY benefactor
ORDER BY name ASC) 
UNION ALL
(SELECT table_realstate.benefactor, name, SUM(rent) unpaid 
FROM table_realstate, table_benefactor 
WHERE flag = 0 
AND table_realstate.benefactor = table_benefactor.id 
AND table_realstate.paidrent = 0
GROUP BY benefactor
ORDER BY name ASC) 
UNION ALL
(SELECT table_realstate.benefactor, name, SUM(rent) paid 
FROM table_realstate, table_benefactor 
WHERE flag = 0 AND table_realstate.benefactor = table_benefactor.id 
AND table_realstate.paidrent = 1
GROUP BY benefactor
ORDER BY name ASC)

结果：

name | totalrent 
_________________
Jhon | 1,000.00  
Doe  | 2,500.00  
Chris| 800.00    
Jhon | 100.00  
Doe  | 500.00 
Jhon | 900.00 
Doe  | 2,000.00 
Chris| 800.00

所以我一直在尝试这样的事情：

如果我这样做，结果的结构就是我需要的，但数学和恩人的名字出错了 - ＆gt;

SELECT table_benefactor.name, SUM(rent) totalrent, SUM(rent) unpaid, SUM(rent) paid
    FROM table_realstate, table_benefactor,
    (SELECT table_realstate.benefactor, name, SUM(rent) totalrent 
    FROM table_realstate, table_benefactor 
    WHERE flag = 0 
    AND table_realstate.benefactor = table_benefactor.id 
    UNION ALL
    SELECT table_realstate.benefactor, name, SUM(rent) unpaid 
    FROM table_realstate, table_benefactor 
    WHERE flag = 0 
    AND table_realstate.benefactor = table_benefactor.id 
    AND table_realstate.paidrent = 0 
    UNION ALL
    SELECT table_realstate.benefactor, name, SUM(rent) paid 
    FROM table_realstate, table_benefactor 
    WHERE flag = 0 AND table_realstate.benefactor = table_benefactor.id 
    AND table_realstate.paidrent = 1) abc 
GROUP BY table_realstate.benefactor

结果：

name | totalrent | paid | unpaid
_________________________________
Jhon | 119.200   |119.200  | 119.200
Jhon | 1,800.02  |1,800.02 | 1,800.02
Jhon | 29,1964.2 |29,1964.2| 29,1964.2
Jhon | 27,000.00 |27,000.00| 27,000.00

正如你所看到的那样，恩人的名字会重演，价值就会到处都是。

我已经尝试了其他的东西，但是不想再发帖了。

Answer 1

我认为你只想要条件聚合。如果paidrent标志仅采用值0和1，如代码所示，那么这很简单：

select b.name, sum(re.rent) as total_rent,
       sum(re.rent * re.paidrent) as paid,
       sum(re.rent * (1 - re.paidrent)) as paid,
from table_benefactor b join
     table_realestate re
     on b.id = re.benefactor
group by b.id, b.name;

Mysql嵌套查询，1个表，3个查询，个别结果

1 个答案: