如何使用postgresql-simple检索JSON jsonb值？

Question

我在postgresql数据库中有一个类型为jsonb的列(jsonExample)。

selectCALogs :: IO [(Int, Object)]
selectCALogs = do
  con <- connection
  query_ con "select \"clusterId\", \"jsonExample\" from cluster"

这会产生以下错误：

    • 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"
   |   ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

我怎样才能让它返回一个JSON对象--使用aeson还是别的什么？

Answer 1

看看这里的FromField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"