如何使用 postgresql-simple 检索 JSON jsonb 值?
How to retrieve JSON jsonb value with postgresql-simple?
我在 postgresql 数据库中有一列 (jsonExample
),类型为 jsonb
。
selectCALogs :: IO [(Int, Object)]
selectCALogs = do
con <- connection
query_ con "select \"clusterId\", \"jsonExample\" from cluster"
这给出了一个错误:
• No instance for (Database.PostgreSQL.Simple.FromField.FromField
(unordered-containers-0.2.10.0:Data.HashMap.Base.HashMap
Text Value))
arising from a use of ‘query_’
• In a stmt of a 'do' block:
query_ con "select \"clusterId\", \"clusterCALogs\"
from cluster"
In the expression:
do con <- connection
query_ con "select \"clusterId\", \"clusterCALogs\"from cluster"
In an equation for ‘selectCALogs’:
selectCALogs
= do con <- connection
query_ con "select \"clusterId\",
\"clusterCALogs\" from cluster"
|
80 | query_ con "select \"clusterId\", \"clusterCALogs\"
from cluster"
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
我怎样才能得到它 return 一个 JSON 对象 - 使用 aeson 或其他东西?
查看此处的 FromField
个实例 (http://hackage.haskell.org/package/postgresql-simple-0.6.2/docs/Database-PostgreSQL-Simple-FromField.html#t:FromField) 我意识到它应该是 Value
而不是 Object
。
因此:
selectCALogs :: IO [(Int, Value)]
selectCALogs = do
con <- connection
query_ con "select \"clusterId\", \"jsonExample\" from cluster"
我在 postgresql 数据库中有一列 (jsonExample
),类型为 jsonb
。
selectCALogs :: IO [(Int, Object)]
selectCALogs = do
con <- connection
query_ con "select \"clusterId\", \"jsonExample\" from cluster"
这给出了一个错误:
• No instance for (Database.PostgreSQL.Simple.FromField.FromField
(unordered-containers-0.2.10.0:Data.HashMap.Base.HashMap
Text Value))
arising from a use of ‘query_’
• In a stmt of a 'do' block:
query_ con "select \"clusterId\", \"clusterCALogs\"
from cluster"
In the expression:
do con <- connection
query_ con "select \"clusterId\", \"clusterCALogs\"from cluster"
In an equation for ‘selectCALogs’:
selectCALogs
= do con <- connection
query_ con "select \"clusterId\",
\"clusterCALogs\" from cluster"
|
80 | query_ con "select \"clusterId\", \"clusterCALogs\"
from cluster"
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
我怎样才能得到它 return 一个 JSON 对象 - 使用 aeson 或其他东西?
查看此处的 FromField
个实例 (http://hackage.haskell.org/package/postgresql-simple-0.6.2/docs/Database-PostgreSQL-Simple-FromField.html#t:FromField) 我意识到它应该是 Value
而不是 Object
。
因此:
selectCALogs :: IO [(Int, Value)]
selectCALogs = do
con <- connection
query_ con "select \"clusterId\", \"jsonExample\" from cluster"