在查询中重用 SUM OVER PARTITION 返回值答案

【问题标题】：Reuse SUM OVER PARTITION return value in query在查询中重用 SUM OVER PARTITION 返回值
【发布时间】：2017-12-09 23:28:09
【问题描述】：

我正在寻找一种方法来避免在查询中执行两次计算，例如：

SELECT DISTINCT
    coins.price_btc,
    coins.price_eur,
    coins.price_usd,
    coins.currency_name,
    coins.currency_symbol,
    SUM ( market_transactions.quantity ) OVER ( PARTITION BY market_transactions.market_coin_id ) * coins.price_eur AS holdings_eur,
    SUM ( market_transactions.quantity ) OVER ( PARTITION BY market_transactions.market_coin_id ) * coins.price_usd AS holdings_usd,
    SUM ( market_transactions.quantity ) OVER ( PARTITION BY market_transactions.market_coin_id ) * coins.price_btc AS holdings_btc,
    SUM ( market_transactions.quantity ) OVER ( PARTITION BY market_transactions.market_coin_id ) AS holdings 
FROM
    market_transactions
    INNER JOIN coins ON coins.id = market_transactions.market_coin_id 
WHERE
    market_transactions.user_id = 1 
ORDER BY
    coins.currency_symbol

我不确定分区总和是否一直在运行。

感谢任何指点，我确信查询也可以优化，但我不确定从哪里开始。

CREATE TABLE "public"."coins" (
  "id" int8 NOT NULL DEFAULT nextval('coins_id_seq'::regclass),
  "currency_symbol" text COLLATE "pg_catalog"."default" NOT NULL DEFAULT NULL,
  "currency_name" text COLLATE "pg_catalog"."default" NOT NULL DEFAULT NULL,
  "price_usd" numeric(16,7) NOT NULL DEFAULT NULL,
  "price_eur" numeric(16,7) NOT NULL DEFAULT NULL,
  "price_btc" numeric(16,7) NOT NULL DEFAULT NULL,
  CONSTRAINT "coins_pkey" PRIMARY KEY ("id")
)

CREATE TABLE "public"."market_transactions" (
  "id" int8 NOT NULL DEFAULT nextval('market_transactions_id_seq'::regclass),
  "user_id" int4 NOT NULL DEFAULT NULL,
  "quantity" numeric(18,8) NOT NULL DEFAULT NULL,
  "market_coin_id" int4 DEFAULT NULL,
  CONSTRAINT "market_transactions_pkey" PRIMARY KEY ("id")
)

一个用户有许多涉及硬币的交易（market_transactions.market_coin_id 是coins.id），我试图将每个交易的拥有数量（market_transactions.quantity）相加，然后将该值乘以表示的硬币价格以不同的货币（btc、eur、usd）

【问题讨论】：

请分享表结构、示例数据、预期输出以及查询尝试生成的结果的描述。作为一般经验法则，每当我看到 SELECT DISTINCT 和 Window 函数时，我都会怀疑查询的逻辑正确性。
感谢您的检查，我添加了简化的表结构，插入需要一些时间，但应该很容易。

标签： sql postgresql

【解决方案1】：

我建议在joining 和做之前聚合：

SELECT c.*,
       mt.quantity * c.price_eur AS holdings_eur,
       mt.quantity * c.price_usd AS holdings_usd,
       mt.quantity * c.price_btc AS holdings_btc,
       mt.quantity * c.market_coin_id AS holdings 
FROM coins c JOIN
     (SELECT mt.market_coin_id, SUM(mt.quantity) as quantity
      FROM market_transactions t
      WHERE mt.user_id = 1 
      GROUP BY mt.market_coin_id
     ) mt
     ON c.id = mt.market_coin_id 
ORDER BY c.currency_symbol

【讨论】：

【解决方案2】：

对查询运行 EXPLAIN（即 EXPLAIN SELECT DISTINCT ...）并查看查询计划是什么。最有可能的是，它只运行一次窗口函数。如果它多次运行，请尝试添加外部 SELECT：

SELECT DISTINCT 
  price_btc, 
  price_eur, 
  price_usd, 
  currency_name,
  currency_symbol,
  holdings * price_eur AS holdings_eur,
  holdings * price_usd AS holdings_usd,
  holdings * price_btc AS holdings_btc,
  holdings
FROM (
    SELECT
    coins.price_btc,
    coins.price_eur,
    coins.price_usd,
    coins.currency_name,
    coins.currency_symbol,
    SUM ( market_transactions.quantity ) OVER ( PARTITION BY market_transactions.market_coin_id ) AS holdings 
FROM
    market_transactions
    INNER JOIN coins ON coins.id = market_transactions.market_coin_id 
WHERE
    market_transactions.user_id = 1 
) src
ORDER BY
    currency_symbol

【讨论】：