Question

我正在寻找一种方法来保存一些计算，使其在查询中保存两次，如：

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")
)

我不确定分区的总和是否一直在运行。

感谢任何指针，我确信查询也可以优化，但我不确定从哪里开始。

market_transactions.market_coin_id

用户有很多涉及硬币的交易（coins.id是market_transactions.quantity），我正在尝试对每个交易的数量（editor.setData()）求和，然后将此值乘以以不同货币表示的硬币价格（英国法郎，欧元，美元）

Answer 1

在查询上运行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

Answer 2

我建议在join之前进行汇总并执行：

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

在查询中重用SUM OVER PARTITION返回值

2 个答案: