Question

我正在尝试从三个SQL表中加入数据。

表格格式如下：

客户端

╔════════╗
║ CLIENT ║
╠════════╣
║ A      ║
║ B      ║
║ C      ║
║ D      ║
╚════════╝

work_times

╔════════╦══════════╦════════╦════════════╗
║ Client ║   Work   ║ Amount ║    Date    ║
╠════════╬══════════╬════════╬════════════╣
║ A      ║ Web Work ║     10 ║ 2013-01-12 ║
║ B      ║ Research ║     20 ║ 2013-01-20 ║
║ A      ║ Web Work ║     15 ║ 2013-01-21 ║
║ C      ║ Research ║     10 ║ 2013-01-28 ║
╚════════╩══════════╩════════╩════════════╝

费用

╔════════╦══════════╦════════╦════════════╗
║ Client ║   Item   ║ Amount ║    Date    ║
╠════════╬══════════╬════════╬════════════╣
║ A      ║ Software ║     10 ║ 2013-01-12 ║
║ B      ║ Software ║     20 ║ 2013-01-20 ║
╚════════╩══════════╩════════╩════════════╝

我想要一个查询，返回每个客户的工作和费用的计数和总和，即：

╔════════╦═══════════╦═══════════╦══════════════╦══════════════╗
║ CLIENT ║ COUNTWORK ║ WORKTOTAL ║ COUNTEXPENSE ║ EXPENSETOTAL ║
╠════════╬═══════════╬═══════════╬══════════════╬══════════════╣
║ A      ║         2 ║        25 ║            1 ║           10 ║
║ B      ║         1 ║        20 ║            1 ║           20 ║
║ C      ║         1 ║        10 ║            0 ║            0 ║
╚════════╩═══════════╩═══════════╩══════════════╩══════════════╝

到目前为止，我有以下内容：

SELECT clients.Client,
 COUNT(distinct work_times.id) AS num_work,
 COUNT(expenses.id) AS num_expenses
FROM
 clients
 INNER JOIN work_times ON work_times.Client = clients.Client
   INNER JOIN expenses ON expenses.Client = work_times.Client
GROUP BY
  clients.Client

这似乎沿着正确的路线，但它跳过了没有费用的客户端，似乎将num_expenses乘以num_work。我还想添加一个WHERE子句来指定只返回两个日期之间的工作时间和费用。我需要对查询进行哪些更改才能获得所需的输出？

Answer 1

您需要单独计算子查询中的值。最外层查询的WHERE子句的目的是过滤掉至少在一个表上记录的记录。因此，在这种情况下，Client D将不会显示在结果列表中。

SELECT  a.*,
        COALESCE(b.totalCount, 0) AS CountWork,
        COALESCE(b.totalAmount, 0) AS WorkTotal,
        COALESCE(c.totalCount, 0) AS CountExpense,
        COALESCE(c.totalAmount, 0) AS ExpenseTotal
FROM    clients A
        LEFT JOIN
        (
            SELECT  Client, 
                    COUNT(*) totalCount,
                    SUM(Amount) totalAmount
            FROM    work_times
            WHERE   DATE BETWEEN '2013-01-01' AND '2013-02-01'
            GROUP   BY Client
        ) b ON a.Client = b.Client
        LEFT JOIN
        (
            SELECT  Client, 
                    COUNT(*) totalCount,
                    SUM(Amount) totalAmount
            FROM    expenses
            WHERE   DATE BETWEEN '2013-01-01' AND '2013-02-01'
            GROUP   BY Client
        ) c ON a.Client = c.Client
WHERE   b.Client IS NOT NULL OR
        c.Client IS NOT NULL

SQLFiddle Demo

更新

╔════════╦═══════════╦═══════════╦══════════════╦══════════════╗
║ CLIENT ║ COUNTWORK ║ WORKTOTAL ║ COUNTEXPENSE ║ EXPENSETOTAL ║
╠════════╬═══════════╬═══════════╬══════════════╬══════════════╣
║ A      ║         2 ║        25 ║            1 ║           10 ║
║ B      ║         1 ║        20 ║            1 ║           20 ║
║ C      ║         1 ║        10 ║            0 ║            0 ║
╚════════╩═══════════╩═══════════╩══════════════╩══════════════╝

Answer 2

您可以将group by移至子查询，这样就不会为每work_time重复expense次{。}}。获得子查询后，可以轻松地向两者添加日期过滤器：

SELECT  clients.Client
,       work_times.cnt AS num_work
,       work_times.total AS total_work
,       expenses.cnt AS num_expenses
,       expenses.total AS total_expenses
FROM    clients
LEFT JOIN
        (
        SELECT  Client
        ,       COUNT(DISTINCT id) as cnt
        ,       SUM(Amount) as total
        FROM    work_times
        WHERE   Date between '2013-01-01' and '2013-02-01'
        GROUP BY
                Client
        ) work_times
ON      work_times.Client = clients.Client
LEFT JOIN
        (
        SELECT  Client
        ,       COUNT(DISTINCT id) as cnt
        ,       SUM(Amount) as total
        FROM    expenses
        WHERE   Date between '2013-01-01' and '2013-02-01'
        GROUP BY
                Client
        ) expenses
ON      expenses.Client = clients.Client

Answer 3

我没有适当的实例在这里进行测试，但我可能会这样开始，然后如果我可以进一步改进查询那么...

select 
  T1.client, 
  ce AS 'Count Work', 
  am AS 'Work Total', 
  ci AS 'Count Expense', 
  am2 AS 'Expense Total' 
from (
  select 
    client, 
    count (work) as ce, 
    sum(amount) as am 
  FROM 
    clients 
      left join work_times 
      on fk_client=client 
  group by 
    fk_client
) T1 
left join (
  select 
    client, 
    count(item) as ci, 
    sum(amount) as am2 
  from 
    clients 
      left join expenses 
      on fk_client=client 
  group by fk_client
) T2 
where T1.client=T2.client;

也许这看起来很复杂，但它确保每个客户只有一行。也许以后它更具可读性......

INNER JOIN查询使用三个表来返回符合条件的行的Count和Sum

3 个答案: