查询生成器：在关联数据上具有多个 SUM 的子查询答案

【问题标题】：Query Builder: Subquery with multiple SUMs on associated data查询生成器：在关联数据上具有多个 SUM 的子查询
【发布时间】：2021-09-09 13:24:14
【问题描述】：

我想就使用查询构建器从关联模型中获取多个总和的方法寻求建议。

一共有三个表：

invoices    invoice_items                               payment_receipts
--------    -------------                               -------------
id | name   id| invoice_id  | invoice_qty   unit_price  id| invoice_id  | receipt_amount
===|======  ==========================================  ================================    
 1 |INV01   1| 1            | 1300          |12.00      1 | 1           | 1000
 2 |INV02   2| 1            | 2600          |9.00       2 | 1           | 2000
 3 |INV03   3| 2            | 1100          |15.00      3 | 3           | 900
            4| 3            | 900           |12:00

对于每张发票，我想要商品总金额的总和（数量 * 价格），以及付款收据的总和。

这个查询（带有子查询）正确地得到了我正在寻找的结果：

SELECT Invoices.id, Invoices.invoice_name, InvoiceItemSum.SumOfAmount, PaymentSum.SumOfPaymentAmount
  FROM Invoices
  INNER JOIN (
    SELECT invoice_id, SUM(Invoice_items.invoice_qty * Invoice_items.unit_price) AS SumOfAmount
      FROM Invoice_items
      GROUP BY Invoice_id
  ) InvoiceItemSum ON InvoiceItemSum.Invoice_id = Invoices.id
  LEFT JOIN (
    SELECT Invoice_id, SUM(Payment_receipts.receipt_amount) AS SumOfPaymentAmount
    FROM Payment_receipts
    GROUP BY Invoice_id
  ) PaymentSum ON PaymentSum.Invoice_id = Invoices.id
WHERE Invoices.invoice_id = 33

我可以直接在我的 CakePhp 应用程序中执行这个查询并获得我需要的结果，所以它就是这样工作的。

不过，我想通过查询构建器获得关于更优雅的 CakePHP 方法的建议。

我试过这个：

        $query = $this->Invoices->find()->contain(['InvoiceItems', 'PaymentReceipts']);
        $query->select([
                'Invoices.id',
                'Invoices.invoice_name',
            ]);
        $query->select([
            'total_inv_amt' => $query->func()->sum('InvoiceItems.invoice_qty * InvoiceItems.unit_price'),
            'total_paid_amt' => $query->func()->sum('PaymentReceipts.receipt_amount')
                ])
                ->innerJoinWith('InvoiceItems')
                ->leftJoinWith('PaymentReceipts')
                ->group(['Invoices.id']);
        $query->where(['Invoices.id' => 33]);

但这会导致通过创建此查询将两个总和加倍：

SELECT 
  Invoices.id AS Invoices__id, 
  Invoices.invoice_name AS Invoices__invoice_name, 
  (
    SUM(
      InvoiceItems.invoice_qty * InvoiceItems.unit_price
    )
  ) AS total_inv_amt, 
  (
    SUM(PaymentReceipts.receipt_amount)
  ) AS total_paid_amt 
FROM 
  invoices Invoices 
  INNER JOIN invoice_items InvoiceItems ON Invoices.id = (InvoiceItems.invoice_id) 
  LEFT JOIN payment_receipts PaymentReceipts ON Invoices.id = (PaymentReceipts.invoice_id) 
WHERE 
  Invoices.id = 33 
GROUP BY 
  Invoices.id

我在文档之后尝试了子查询，但结果无数。我也玩过连接，但仍然没有骰子。

我的问题是：使用查询生成器编写此查询的好方法是什么？

提前感谢您的任何建议！

【问题讨论】：

你是不是总结错了？
没错……总和翻了一番。
让我更正一下。如果只有一个发票项目，则结果是正确的。如果有多个发票项目，则总发票金额为 SUM * 2，总付款金额为 SUM * 发票项目数。
请注意，单独运行 InvoiceItems SUM 或 PaymentReceipts SUM 都可以正常工作。当我们将两个连接查询放在一起时，它会乘以总数。

标签： cakephp cakephp-4.x

【解决方案1】：

您可以通过 cakephp JOIN 轻松完成。检查此解决方案它可能会对您有所帮助。直接用编辑器编写，不保证无语法错误！

$this->Invoices->find()
   ->select([
               'Invoices.id',
               'Invoices.invoice_name',
               'total_inv_amt' => $query->func()->sum('InvoiceItems.invoice_qty * InvoiceItems.unit_price'),
               'total_paid_amt' => $query->func()->sum('PaymentReceipts.receipt_amount')
    ])
    ->join([
            'table' => 'InvoiceItems',
            'alias' => 'InvoiceItems',
            'type' => 'Inner',
            'conditions' => [
                 // your condition for InvoiceItems
             ],
    ])
    ->join([
            'table' => 'PaymentReceipts',
            'alias' => 'PaymentReceipts',
            'type' => 'LEFT',
            'conditions' => [
                // condition for PaymentReceipts
            ],
    ])

【讨论】：

感谢您的回复。实际上，这种方法（加上 ->group(['Invoices.id'])）会为每张发票获得一个结果，但两者的 SUM 是一个巨大的总数，并且每个 invoice.id 的值相同。跨度>
基于有效的原始 SQL，我们需要在 JOINS 中获取一个 SELECT 作为子查询......我认为。

【解决方案2】：

好的，最后在查看了 Alimon 的答案以及关于子查询（例如 this 和 this）的其他几个问题之后，我已经找到了正确的 Query Builder 解决方案。这里是：

        $subquery_a = $this->Invoices->InvoiceItems->find('all');
        $subquery_a
            ->select(['totalinvoiceamt' => $subquery_a->func()->sum('invoice_qty * unit_price') ])
            ->where([
                'InvoiceItems.invoice_id = Invoices.id'
            ]);
    
        $subquery_b = $this->Invoices->PaymentReceipts->find('all');
        $subquery_b
            ->select(['totalpaymentamt' => $subquery_b->func()->sum('receipt_amount') ])
            ->where([
                'PaymentReceipts.invoice_id = Invoices.id'
            ]);

        $query = $this->Invoices->find('all')
            ->select([
            'Invoices.id',
            'Invoices.invoice_name',
            'InvoiceItems__total_invoice_amount' => $subquery_a,
            'PaymentReceipts__total_payments_amount' => $subquery_b
            ])
            ->join([
                [
                    'table'     => 'invoice_items',
                    'alias'     => 'InvoiceItems',
                    'type'      => 'INNER',
                    'conditions'=> [
                        'Invoices.id = InvoiceItems.invoice_id'
                    ]
                ]
            ])
            ->join([
                [
                    'table'     => 'payment_receipts',
                    'alias'     => 'PaymentReceipts',
                    'type'      => 'LEFT',
                    'conditions'=> [
                        'Invoices.id = PaymentReceipts.invoice_id'
                    ]
                ]
            ])
            ->group('InvoiceItems.invoice_id');
            $query->where(['Invoices.id' => 33]);

结果和直接查询一样，虽然SQL看起来和手动的有点不一样，但是结果是一样的。

感谢 Alimon 等人的帮助。

【讨论】：