我正在寻找一种方法来避免在查询中进行两次计算,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
我不确定分区上的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")
)
用户有很多涉及硬币的交易(market_transactions.market_coin_id
is coins.id
),我正在尝试对拥有的数量求和(market_transactions.数量
),然后将这个值乘以不同货币(btc,eur,usd)表示的硬币价格
最佳答案
我建议在join
ing 和做之前进行聚合:
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
关于sql - 在查询中重用 SUM OVER PARTITION 返回值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47734327/