Question

我对以下相当简单的设置有困难：

CREATE TABLE IF NOT EXISTS invoices (
id int(11) NOT NULL auto_increment,
PRIMARY KEY (id)
);

CREATE TABLE IF NOT EXISTS invoices_items (
id int(11) NOT NULL auto_increment,
invoice_id int(11) NOT NULL,
description text NOT NULL,
amount decimal(10,2) NOT NULL default '0.00',
PRIMARY KEY (id)
);

CREATE TABLE IF NOT EXISTS invoices_payments (
id int(11) NOT NULL auto_increment,
invoice_id int(11) NOT NULL,
amount decimal(10,2) NOT NULL default '0.00',
PRIMARY KEY (id)
);

一些数据：

INSERT INTO invoices (id) VALUES (1);

INSERT INTO invoices_items (id, invoice_id, description, amount) VALUES
(1, 1, 'Item 1', '750.00'),
(2, 1, 'Item 2', '750.00'),
(3, 1, 'Item 3', '50.00'),
(4, 1, 'Item 4', '150.00');

INSERT INTO invoices_payments (id, invoice_id, amount) VALUES
(1, 1, '50.00'),
(2, 1, '1650.00');

和sql产生了不寻常的结果：

select  invoices.id,
        ifnull(sum(invoices_payments.amount),0) as payments_total, 
        ifnull(count(invoices_items.id),0) as item_count 
from    invoices
            left join invoices_items on invoices_items.invoice_id=invoices.id 
            left join invoices_payments on invoices_payments.invoice_id=invoices.id 
group by invoices.id

导致（错误的）输出

id  payments_total  item_count
1   6800.00     8

现在，事实证明只有四个'invoice_item'行，我不明白为什么mysql没有正确分组。

修改

我知道我可以这样做：

select x.*, ifnull(sum(invoices_payments.amount),0) as payments_total from (
    select  invoices.id,
            ifnull(count(invoices_items.id),0) as item_count 
    from    invoices
                left join invoices_items on invoices_items.invoice_id=invoices.id 
    group by invoices.id
) as x left join invoices_payments on invoices_payments.invoice_id=x.id
group by x.id

但我想知道我是否在第一个查询中做错了什么 - 我无法立即看到为什么第一个查询给出了错误的结果！：（

Answer 1

您的联接逻辑不正确。在您的联接中，您指定invoices_items.invoice_id = invoices.id。您还可以指定invoices_payments.invoice_id = invoices.id。由于传递性，你最终得到：

invoices_items.invoice_id    = invoices.id
invoices_payments.invoice_id = invoices.id
invoice_items.invoice_id     = invoices_payments.invoice_id

2张发票付款的总和为1700美元。对于每个发票付款，有4个invoice_items满足上述关系。 $ 1700 * 4 = $ 6800。

对于每个发票项目，将有两个发票付款满足上述关系。 4个发票项目* 2 = 8个计数。

Answer 2

编辑：

啊，对不起 - 现在看你的观点。您获得意外结果的原因是此查询：

SELECT * 
FROM invoices
    LEFT JOIN invoices_items ON invoices_items.invoice_id = invoices.id
    LEFT JOIN invoices_payments ON invoices_payments.invoice_id = invoices.id;

结果如下：

id  id invoice_id   description amount  id  invoice_id  amount
1    1  1            Item 1      750.00  1   1           50.00
1    1  1            Item 1      750.00  2   1           1650.00
1    2  1            Item 2      750.00  1   1           50.00
1    2  1            Item 2      750.00  2   1           1650.00
1    3  1            Item 3      50.00   1   1           50.00
1    3  1            Item 3      50.00   2   1           1650.00
1    4  1            Item 4      150.00  1   1           50.00
1    4  1            Item 4      150.00  2   1           1650.00

正如您所看到的，每个invoices_items记录每次都会获得invoices_payments条记录。你将不得不分别抓住它们（即分组）。

请注意，初始查询中的GROUP BY子句是多余的。

以下是您的需求：

SELECT
    invoices.id,
    payments_total.payments_total, 
    IFNULL(COUNT(invoices_items.id),0) AS item_count 
FROM invoices
    LEFT JOIN invoices_items ON invoices.id = invoices_items.invoice_id
    LEFT JOIN (
        SELECT invoice_id,
            IFNULL(SUM(invoices_payments.amount),0) AS payments_total
            FROM invoices_payments
            GROUP BY invoice_id
        ) AS payments_total ON invoices.id  = payments_total.invoice_id
;

Answer 3

有两张表有很多：一张与发票的关系。你的数量是笛卡尔积。

付款应该应用于发票，而不是发票项目。首先获取发票总额，然后加入付款。

这可能类似于您所寻找的内容：

SELECT
    invoice_total.invoice_id,
    invoice_total.amount as invoice_amount,
    payments_total.amount as total_paid
FROM
    (
        SELECT
            invoice_id,
            SUM(amount) as amount
        FROM
            invoices_items
        GROUP BY
                invoice_id
        ) invoice_total
    INNER JOIN
            (
            SELECT 
                invoice_id,
                SUM(amount) as amount
            FROM
                invoices_payments
            GROUP BY
                invoice_id
            ) payments_total
    ON invoice_total.invoice_id  = payments_total.invoice_id;

通过提供不寻常的结果，mysql多对一和分组

修改

3 个答案: