Question

data CumulativeRevenue = CumulativeRevenue
  { payment_date :: T.Text
  , amount       :: Double
  , sum          :: Double
  } deriving (Show, Generic, Aeson.ToJSON, Aeson.FromJSON)

instance Postgres.FromRow CumulativeRevenue where
  fromRow = CumulativeRevenue
            <$> Postgres.field
            <*> Postgres.field
            <*> Postgres.field

cumulativeRevenue :: Postgres.Connection -> IO [CumulativeRevenue]
cumulativeRevenue conn = Postgres.query_ conn
  "SELECT payment_date, amount, sum(amount) OVER (ORDER by payment_date) \
  \ FROM (\
  \ SELECT CAST (payment_date as TEXT) AS payment_date, SUM(amount) AS \
  \ amount \
  \ FROM payment \
  \ GROUP BY CAST(payment_date AS TEXT) \
  \ ) p \
  \ ORDER BY payment_date \
  \"

目前，我有上面的代码。完整的代码是here。累积收入给出了以下异常。您可以忽略spock部分。

Spock Error while handling ["cumulative"]: Incompatible {errSQLType = "numeric", errSQLTableOid = Nothing, errSQLField = "amount", errHaskellType = "Double", errMessage = "types incompatible"}

我不清楚在CumulativeRevenue中为数量和总和字段指定什么。有人可以帮我吗？使用postgres-simple库时，有没有更简单的方法来找出从Haskell类型到SQL类型的类型转换，反之亦然？

Answer 1

FromField有一个Double实例，但是它不适用于numeric，因为numeric是精度为as per the PostgresSQL documentation的变量。如果查看postgresql-simple documentation，您会发现Ratio Integer（又名Rational）实例确实支持numeric。

所以您可以在这里做两件事之一：

在Rational字段中使用Double代替amount
确定您对丢失精度没事（尽管您为什么要在SQL端使用numeric？），然后做类似的事情

import Data.Ratio

loseNumericPrecision :: Postgres.RowParser Double
loseNumericPrecision = fmap fromRational Postgres.field

，并将其代替Postgres.field用于与numeric值相对应的字段。

Postgres-simple的FromRow实例定义错误

1 个答案: