Yesod Single RawSql並將其作爲JSON格式返回

Question

希望您能幫助我解決這個問題。

我有一個rawSql：

latestPrice :: Handler [(Single PricesId, Single PersistValue, Single PersistValue, Single PersistValue, Single PersistValue, Single PersistValue)] 
latestPrice = do 
    runDB $ rawSql qryStr [] 
     where qryStr = " SELECT prices.id,\ 
          \CONCAT(cities.name, '(', states.code, '), ', countries.name) as city, \ 
          \prices.s_id, \ 
          \prices.s_name, \ 
          \prices.quality, \ 
          \AVG(prices.amount) as amount \ 
        \FROM prices \ 
        \INNER JOIN cities ON cities.id = prices.city_id \ 
        \INNER JOIN states ON states.id = cities.state_id \ 
        \INNER JOIN countries ON countries.id = states.country_id \ 
        \WHERE prices.city_id > 0 \ 
        \GROUP BY prices.id, city, prices.strain_id, prices.strain_name, prices.quality, prices.quality;"; 

，我必須調用該函數應該返回一個JSON格式的處理程序：

getLatestPriceSubmissionR:: Handler (Value) 
getLatestPriceSubmissionR = do 
    results <- latestPrice 
    return map (
    \(Single id, Single city, Single s_name, Single strain_id, Single quality, Single amount) -> [ "id" .= id,"city" .= city,"s_id" .= s_id,"s_name" .= s_name, "quality" .= quality,"amount" .= amount ] 
    ) results 

我的問題是： 做我的代碼在處理器getLatestPriceSubmissionR是正確的？ 它確實給我不僅沒有警告消息的錯誤消息：

Couldn't match type [[aeson-0.8.0.0:Data.Aeson.Types.Internal.Pair]] 
      with Value 
Expected type: HandlerT App IO Value 
Actual type: HandlerT 
      App IO [[aeson-0.8.0.0:Data.Aeson.Types.Internal.Pair]] … 

也

No instance for (ToTypedContent [Value]) 
arising from a use of yesodRunner 

希望大家幫幫我。

預先感謝您

Answer 1

線

return map (\(...) -> [ ... ]) results 

看起來不對，你可能意味着

return $ map (\(...) -> [ ... ]) results 

而且，這將返回一個列表的列表，這是不是一個Value在類型簽名。

而且，這是一個錯誤，而不是一個警告：

Couldn't match type [[aeson-0.8.0.0:Data.Aeson.Types.Internal.Pair]] 
      with Value 
Expected type: HandlerT App IO Value 
Actual type: HandlerT 
      App IO [[aeson-0.8.0.0:Data.Aeson.Types.Internal.Pair]] …