MySQL中SUM的多行總和

Question

我有一個查詢，返回每個order的total列：

SELECT o.id, o.created, o.status, o.shipping,
SUM(op.price*op.amount*(1+op.tax/100.0))+COALESCE(o.shipping, 0)*(1+o.shipping_tax/100.0)+COALESCE(o.payment_fee, 0)*(1+o.shipping_tax/100.0) AS total

FROM orders AS o
LEFT OUTER JOIN order_products as op ON o.id=op.order_id
GROUP BY o.id;

查詢的輸出

現在，我要計算總收入，因此將是這些total列的total 。

我嘗試使用像這樣的子查詢：

SELECT 
SUM(total) FROM (
    SELECT
    SUM(op.price*op.amount*(1+op.tax/100.0))+COALESCE(o.shipping, 0)*(1+o.shipping_tax/100.0)+COALESCE(o.payment_fee, 0)*(1+o.shipping_tax/100.0) AS total
    FROM orders AS o
    LEFT OUTER JOIN order_products as op ON o.id=op.order_id
    GROUP BY o.id
) AS foo
;

查詢的輸出

然而：

• 結果列名為SUM(total) ，而不是foo
  • 似乎AS foo用於子查詢，而不是頂級SUM()
• 沒有AS foo ，它將無法正常工作
• _{該查詢返回的結果可能不正確， 但這可能是由於rounding引起的 。}

查詢以求和SUM的多行，應該如何看待？

注意： 這些NULL總數 幾乎肯定不會在生產中存在，但是最好使用 COALESCE(x, 0)來確保查詢正確執行。 似乎這樣的SUM（）可以使用NULL值嗎？

Answer 1

如果您想為名稱加上別名，則必須在選擇之前而不是之后進行設置，並且要檢查是否為null，必須為此設置一個值：

  SELECT 
    SUM(COALESCE(a.total,0)) as foo FROM (
        SELECT
        SUM(op.price*op.amount*(1+op.tax/100.0))+COALESCE(o.shipping, 0)*(1+o.shipping_tax/100.0)+COALESCE(o.payment_fee, 0)*(1+o.shipping_tax/100.0) AS total
        FROM orders AS o
        LEFT OUTER JOIN order_products as op ON o.id=op.order_id
        GROUP BY o.id
    ) AS a

Answer 2

SELECT
  SUM(
     COALESCE(op.order_total, 0)
    +COALESCE(o.shipping, 0)*(1+o.shipping_tax/100.0)
    +COALESCE(o.payment_fee, 0)*(1+o.shipping_tax/100.0)
  )
    AS grand_total
FROM
  orders   AS o
LEFT OUTER JOIN
(
  SELECT
    order_id,
    SUM(price*amount*(1+tax/100.0))  AS order_total
  FROM
    order_products
  GROUP BY
    order_id
)
  AS op
    ON o.id=op.order_id

編輯：

你說過：
SUM(x+y+z) / COUNT(op.order_total)為506
AVG(x+y+z)得到532

在我看來，這意味着當x NOT NULL為NULL時，某些x+y+z NOT NULL ，因此看來y或z為NULL 。

由於y和z為COALESCE(?, 0) * (1+o.shipping_tax/100.0) ，因此感覺 shipping_tax有時為NULL 。

試試這個查詢...

SELECT
  SUM(
     COALESCE(op.order_total, 0)
    +COALESCE(o.shipping, 0)*(1+o.shipping_tax/100.0)
    +COALESCE(o.payment_fee, 0)*(1+o.shipping_tax/100.0)
  )
    AS grand_total,
  AVG(
     COALESCE(op.order_total, 0)
    +COALESCE(o.shipping, 0)*(1+o.shipping_tax/100.0)
    +COALESCE(o.payment_fee, 0)*(1+o.shipping_tax/100.0)
  )
    AS grand_average,
  COUNT(
     COALESCE(op.order_total, 0)
    +COALESCE(o.shipping, 0)*(1+o.shipping_tax/100.0)
    +COALESCE(o.payment_fee, 0)*(1+o.shipping_tax/100.0)
  )
    AS grand_row_count,
  COUNT(
    *
  )
    AS set_row_count,
  COUNT(
    o.shipping_tax
  )
    AS shipping_tax_row_count,
  AVG(
     COALESCE(op.order_total, 0)
    +(COALESCE(o.shipping, 0)+COALESCE(o.payment_fee, 0))
     *COALESCE(1+o.shipping_tax/100.0, 1)
  )
    AS revised_grand_average
FROM
  orders   AS o
LEFT OUTER JOIN
(
  SELECT
    order_id,
    SUM(price*amount*(1+tax/100.0))  AS order_total
  FROM
    order_products
  GROUP BY
    order_id
)
  AS op
    ON o.id=op.order_id

我希望 ...
grand_average == grand_total / grand_row_count
set_row_count > grand_row_count
grand_row_count == shipping_tax_row_count

如果是這樣，那么revised_grand_average應該對您有用。

如果不是，那么希望這給您一個開始調查的地方。

Answer 3

您不需要子查詢。 只需從原始查詢中刪除GROUP BY ：

SELECT SUM(op.price*op.amount*(1+op.tax/100.0))+COALESCE(o.shipping, 0)*(1+o.shipping_tax/100.0)+COALESCE(o.payment_fee, 0)*(1+o.shipping_tax/100.0) AS total
FROM orders o LEFT OUTER JOIN
     order_products op
     ON o.id = op.order_id;

這將對所有行進行計算。 它應該快於兩個級別的聚合。

結果將在稱為total的列中。