Question

我有以下查询尝试生成所有客户的列表，他们的项目，每个项目花费的时间，总发票金额和发票总数。

SELECT
    c.name as client,
    p.name as project,
    SUM(t.hours) as hours,
    SUM(ci.amount) as invoice_amount,
    COUNT(ci.id) as number_invoices
FROM
    projects p
    INNER JOIN client_master c ON c.harvest_id = p.client_id
    INNER JOIN timesheet t ON t.project_id = p.id 
    LEFT OUTER JOIN projects_contracts pc ON pc.project_id = p.id
    INNER JOIN contracts_invoices ci ON ci.contract_id = pc.contract_id
GROUP BY
    c.name,
    p.name
ORDER BY
    c.name, 
    p.name,
    t.hours

当我运行查询时，＆＃34;小时＆＃34;正确计算，但invoice_amount和number_invoices的总和/计数不是在上下文中加入项目，而是与表中的发票总数相关联。如果我注释掉其中一个sum（）＆＃39; s，那么我得到了正确的值，但是当它们都进入时都没有。

预期结果

客户A，项目1,1000，$ 10,000,2

客户A，项目2,5000，$ 5,000,1

客户B，项目1,100，$ 3000,1

实际结果

客户A，项目1,1000，$ 150,000,20

客户A，项目2,5000，$ 50,000，81

客户B，项目1,100，$ 83000,12

我已经尝试弄乱了分组和加入声明，但没有找到成功。有什么简单的我可以忽略或者我需要完全重写吗？

Answer 1

由于发票取决于合同，合同是可选的（左连接），使该连接也成为左连接。

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Answer 2

这可能不是最好的方式，但我敢打赌它有效

SELECT
    c.name as client,
    p.name as project,
    t.hours as hours,
    ci.amount as invoice_amount,
    ci.count as number_invoices
FROM
    projects p
    INNER JOIN client_master c 
        ON c.harvest_id = p.client_id
    INNER JOIN (SELECT project_id, SUM(hours) AS hours FROM timesheet GROUP BY project_id) t
        ON t.project_id = p.id 
    LEFT OUTER JOIN projects_contracts pc 
        ON pc.project_id = p.id
    LEFT OUTER JOIN (SELECT contract_id, SUM(amount) AS amount, COUNT(DISTINCT id) AS count FROM contracts_invoices GROUP BY contract_id) ci
        ON ci.contract_id = pc.contract_id
GROUP BY
    c.name,
    p.name
ORDER BY
    c.name, 
    p.name,
    t.hours

MySQL查询结合多个表的总和

2 个答案: