oracle ora-00904与symfony2和doctrine2

时间:2014-06-12 18:48:16

标签: sql oracle symfony doctrine-orm

我有一个转换成这个SQL的学说DQL:

SELECT o0_.id AS ID0, 
o0_.pmn AS PMN1, 
o0_.neteo AS NETEO2, 
sum(o1_.monto_total) AS SCLR3, 
o1_.periodo AS PERIODO4, 
o1_.pagado AS PAGADO5 
FROM operador o0_ 
LEFT JOIN operador_hub o2_ ON o0_.id = o2_.operador_id 
LEFT JOIN outcollect o1_ ON o0_.id = o1_.operador_id 
WHERE o0_.ishub = 0 
AND o0_.neteo = 1 
AND o0_.incluirReportes = 1 
AND o0_.pmn NOT IN ('CHLTM', 'CHLCM') 
AND o0_.id NOT IN (SELECT o3_.id FROM operador o3_ LEFT JOIN operador_hub o4_ ON o3_.id = o4_.operador_id LEFT JOIN outcollect o5_ ON o3_.id = o5_.operador_id WHERE (o4_.desde <= o5_.periodo AND (o4_.hasta >= o5_.periodo OR o4_.hasta IS NULL))) 
AND o0_.activo = 1 
GROUP BY o0_.pmn, PERIODO4 
ORDER BY o0_.pmn ASC

这会产生ORA-00904: "PERIODO4": identificador no válido

该错误与GROUP BY o0_.pmn,PERIODO4行

直接相关

似乎与不能将查询转换为ORACLE语法的学说有关。我正在使用OCI8驱动程序。

有什么想法吗?

----编辑----

这是DQL:

SELECT partial operador.{id, pmn, neteo},                           
sum(outcollect.montoTotal) as montoOut,
outcollect.periodo as periodo,
outcollect.pagado as pagado
FROM RoamingOperadoresBundle:Operador operador
LEFT JOIN operador.operadorHub operadorHub 
LEFT JOIN operador.outcollect outcollect
WHERE operador.ishub = 0
AND operador.neteo = 1                          
AND operador.incluirReportes = 1
AND operador.pmn not in('CHLTM', 'CHLCM')
AND operador.id not in(SELECT op.id 
FROM RoamingOperadoresBundle:Operador op
LEFT JOIN op.operadorHub opHub
LEFT JOIN op.outcollect out
WHERE (opHub.desde <= out.periodo and (opHub.hasta >= out.periodo or opHub.hasta is null)))
AND operador.activo = 1
GROUP BY operador.pmn, periodo ORDER BY operador.pmn

1 个答案:

答案 0 :(得分:1)

我不确定您是如何生成此表表达式的,但是您收到错误是因为您没有在GROUP BY子句中包含所有非聚合列。

另请注意 - 您不能按别名分组。您必须通过在table.column表单中指定该列来引用该列。

它应该是这样的:

GROUP BY o0_.pmn, o0_.neteo, o1_.periodo, o1_.pagado