Question

我的表格结构如下：

units {id, owner_id}
contracts {id, rent, unit_id}
rents {paid, contract_id}

我需要的是两件事的总和：所有rents.paid和contracts.rent的总和。

我的查询是：

SELECT 
   Sum(Rent.paid)      AS Rent__summed_rents
FROM
   rents AS Rent
LEFT JOIN contracts
   ON contracts.id = Rent.id
LEFT JOIN units
   ON contracts.unit_id = units.id AND units.owner_id = 29

我尝试添加contracts.rent * Rent.paid AS Rent__summed_expected_rents，但它没有用。

修改

每个contract都有一列rent，其中包含每个月预期的付款每个rent记录都有一列paid，其中包含实际付款查询应该获得所有付费记录并将它们相加（summed_rents），它还应该告诉我预期收入是多少（contracts.rent * rents.paid）

*编辑＃2 *

rents data:
id | paid | contract_id
 1 | 200  |      6
 1 | 300  |      6
 1 | 500  |      4
 1 | 200  |      4

contracts data:
id | rent |  unit_id
 6 | 500  |    22
 4 | 600  |    22

units data:
id | owner_id
22 |    29

我的期望是：总计租金= 1,200（所有已付款的总和）预期租金= 2,200 =（500 * 2）+（600 * 2）（contract.rent的总和，每个rents记录一个）

Answer 1

以下是我构建查询的方法：

从拥有者的单位开始。很简单：

SELECT u.*
  FROM units u
 WHERE u.owner_id = 29

下一步，获取这些单位的合同：

SELECT c.*
     , u.*
  FROM units u
  JOIN contracts c
    ON c.unit_id = u.id
 WHERE u.owner_id = 29

下一步，获得这些合同的租金

SELECT r.*
     , c.*
     , u.*
  FROM units u
  JOIN contracts c
    ON c.unit_id = u.id
  JOIN rents r
    ON r.contract_id = c.id
 WHERE u.owner_id = 29

- 通过合约ID汇总该结果的行（根据规范，我们需要按合约编号从租金计算，乘以合约租金金额）

SELECT SUM(r.paid) AS total_paid
     , SUM(c.rent) AS total_rent
  FROM units u
  JOIN contracts c
    ON c.unit_id = u.id
  JOIN rents r
    ON r.contract_id = c.id
 WHERE u.owner_id = 29
 GROUP BY c.id

- 通过将结果包装为内联视图并汇总total_paid和总租金来获得总计

SELECT SUM(t.total_paid) AS total_paid
     , SUM(t.total_rent) AS total_rent
  FROM (
         SELECT SUM(r.paid) AS total_paid
              , SUM(c.rent) AS total_rent
           FROM units u
           JOIN contracts c
             ON c.unit_id = u.id
           JOIN rents r
             ON r.contract_id = c.id
          WHERE u.owner_id = 29
          GROUP BY c.id
       ) t

注意：计算给定合约的rents中的行数似乎是获得租金乘数的奇怪方法。但如果这是规范，我们会确保我们的代码能够做到这一点。

我们只需要外部联接来产生“零”付费和“零”租金。如果我们向租户表引入LEFT JOIN，那么我们需要调整表达式以获得total_rent。我们想要使用类似的东西：

   c.rent*COUNT(r.contract_id) AS total_rent

这样的事情：

SELECT IFNULL(SUM(t.total_paid),0) AS total_paid
     , IFNULL(SUM(t.total_rent),0) AS total_rent
  FROM (
         SELECT SUM(r.paid) AS total_paid
              , c.rent*COUNT(r.contract_id) AS total_rent
           FROM units u
           JOIN contracts c
             ON c.unit_id = u.id
           LEFT
           JOIN rents r
             ON r.contract_id = c.id
          WHERE u.owner_id = 29
          GROUP BY c.id
       ) t

Answer 2

这很简单：

select sum(r.paid) as rent_paid,
       sum(c.rent) as expected_rent
  from rents r
  join contracts c on (r.contract_id = c.id)
  join units u on (c.unit_id = u.id)
 where u.owner_id = 29;

见sqlfiddle

MySQL - 在混合列上与SUM（）连接

2 个答案: