多表发票SUM比较

问题描述 投票:11回答:2

假设我在rails应用程序中有3个表:

发票

id  | customer_id  | employee_id  | notes
---------------------------------------------------------------
1   | 1            | 5            | An order with 2 items.
2   | 12           | 5            | An order with 1 item.
3   | 17           | 12           | An empty order.
4   | 17           | 12           | A brand new order.

invoice_items

id  | invoice_id  | price  | name
---------------------------------------------------------
1   | 1           | 5.35   | widget
2   | 1           | 7.25   | thingy
3   | 2           | 1.25   | smaller thingy
4   | 2           | 1.25   | another smaller thingy

invoice_payments

id  | invoice_id  | amount  | method      | notes
---------------------------------------------------------
1   | 1           | 4.85    | credit card | Not enough 
2   | 1           | 1.25    | credit card | Still not enough
3   | 2           | 1.25    | check       | Paid in full

这代表4个订单:

第一项有2项,总计12.60项。它有两笔付款,总付款金额为6.10。此订单已部分支付。

第二项只有一项,一项付款,总计1.25。此订单全额支​​付。

第三个订单没有物品或付款。这对我们很重要,有时我们会使用这种情况。它被认为是全额付款。

最终订单再次有一个项目,共计1.25,但尚未付款。

现在我需要一个查询:

告诉我所有未完全支付的订单;也就是说,所有订单都使得项目总数大于付款总额。

我可以在纯sql中做到这一点:

SELECT      invoices.*,
            invoice_payment_amounts.amount_paid AS amount_paid,
            invoice_item_amounts.total_amount AS total_amount
FROM        invoices
LEFT JOIN   (
                SELECT      invoices.id AS invoice_id,
                            COALESCE(SUM(invoice_payments.amount), 0) AS amount_paid
                FROM        invoices
                LEFT JOIN   invoice_payments
                ON          invoices.id = invoice_payments.invoice_id
                GROUP BY    invoices.id
            ) AS invoice_payment_amounts
ON          invoices.id = invoice_payment_amounts.invoice_id
LEFT JOIN   (
                SELECT      invoices.id AS invoice_id,
                            COALESCE(SUM(invoice_items.item_price), 0) AS total_amount
                FROM        invoices
                LEFT JOIN   invoice_items
                ON          invoices.id = invoice_items.invoice_id
                GROUP BY    invoices.id
            ) AS invoice_item_amounts
ON          invoices.id = invoice_item_amounts.invoice_id
WHERE       amount_paid < total_amount

但是......现在我需要把它变成rails(可能作为范围)。我可以使用find_by_sql,但然后返回一个数组,而不是ActiveRecord :: Relation,这不是我需要的,因为我想将它与其他范围链接(例如,有一个过期的范围,它使用这个)等

所以原始SQL可能不是正确的方式去这里.....但是什么是?我无法在activerecord的查询语言中执行此操作。

我到目前为止最接近的是:

Invoice.select('invoices.*, SUM(invoice_items.price) AS total, SUM(invoice_payments.amount) AS amount_paid').
  joins(:invoice_payments, :invoice_items).
  group('invoices.id').
  where('amount_paid < total')

但是失败了,因为在#1这样的订单上,多次付款,它错误地使订单的价格加倍(由于多次加入),显示它仍然是无偿的。我在SQL中遇到了同样的问题,这就是我按照我的方式构建它的原因。

这有什么想法?

sql ruby-on-rails ruby-on-rails-4
2个回答
9
投票

您可以使用MySQL的group byhaving子句获得结果:

纯MySQL查询:

  SELECT `invoices`.* FROM `invoices` 
  INNER JOIN `invoice_items` ON 
    `invoice_items`.`invoice_id` = `invoices`.`id` 
  INNER JOIN `invoice_payments` ON 
    `invoice_payments`.`invoice_id` = `invoices`.`id` 
  GROUP BY invoices.id 
    HAVING sum(invoice_items.price) < sum(invoice_payments.amount) 

ActiveRecord查询:

Invoice.joins(:invoice_items, :invoice_payments).group("invoices.id").having("sum(invoice_items.price) < sum(:invoice_payments.amount)")

1
投票

在Rails中构建更复杂的查询时,通常Arel Really Exasperates Logicians会派上用场

Arel是Ruby的SQL AST管理器。它

  1. 简化了复杂SQL查询的生成,以及
  2. 适应各种RDBMS。

以下是基于要求如何实现Arel实现的示例

invoice_table = Invoice.arel_table

# Define invoice_payment_amounts
payment_arel_table = InvoicePayment.arel_table
invoice_payment_amounts = Arel::Table.new(:invoice_payment_amounts)
payment_cte = Arel::Nodes::As.new(
  invoice_payment_amounts,
  payment_arel_table
    .project(payment_arel_table[:invoice_id],
             payment_arel_table[:amount].sum.as("amount_paid"))
    .group(payment_arel_table[:invoice_id])
)

# Define invoice_item_amounts
item_arel_table = InvoiceItem.arel_table
invoice_item_amounts =  Arel::Table.new(:invoice_item_amounts)
item_cte = Arel::Nodes::As.new(
  invoice_item_amounts,
  item_arel_table
    .project(item_arel_table[:invoice_id],
             item_arel_table[:price].sum.as("total"))
    .group(item_arel_table[:invoice_id])
)

# Define main query
query = invoice_table
          .project(
            invoice_table[Arel.sql('*')],
            invoice_payment_amounts[:amount_paid],
            invoice_item_amounts[:total]
          )
          .join(invoice_payment_amounts).on(
            invoice_table[:id].eq(invoice_payment_amounts[:invoice_id])
          )
          .join(invoice_item_amounts).on(
            invoice_table[:id].eq(invoice_item_amounts[:invoice_id])
          )
          .where(invoice_item_amounts[:total].gt(invoice_payment_amounts[:amount_paid]))
          .with(payment_cte, item_cte)


res = Invoice.find_by_sql(query.to_sql)
for r in res do
  puts "---- Invoice #{r.id} -----"
  p r
  puts "total: #{r[:total]}"
  puts "amount_paid: #{r[:amount_paid]}"
  puts "----"
end

这将使用您提供给问题的示例数据返回与SQL查询相同的输出。输出:

 <Invoice id: 2, notes: "An order with 1 items.", created_at: "2017-12-18 21:15:47", updated_at: "2017-12-18 21:15:47">
 total: 2.5
 amount_paid: 1.25
 ----
 ---- Invoice 1 -----
 <Invoice id: 1, notes: "An order with 2 items.", created_at: "2017-12-18 21:15:47", updated_at: "2017-12-18 21:15:47">
 total: 12.6
 amount_paid: 6.1
 ----

Arel非常灵活,因此您可以将其作为基础,并根据您可能具有的更具体要求优化查询条件。

我强烈建议您考虑在Invoice表中创建一个缓存列(total,amount_paid)并维护它们,以便您可以避免这种复杂的查询。至少total附加列可以非常简单地创建和填充数据。

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