Question

我有三张桌子，结构如下：

http://dl.dropbox.com/u/2586403/ORMIssues/TableLayout.png

我正在处理的三个对象是：

http://dl.dropbox.com/u/2586403/ORMIssues/Objects.zip

我需要能够获取PartObject，然后提取其所有属性，按Types表中的AttributeName排序。以下是我遇到的问题：

我不能通过它对PartObject中的Attributes属性进行排序 Attribute.AttributeName属性
我无法将Attribute.AttributeName属性添加到ObjectAttribute实体，因为我收到有关列名的错误。 Hibernate将ID放在连接的错误一侧

这是显示错误查询的hibernate日志文件

10/14 16:36:39 [jrpp-12] HIBERNATE DEBUG - select objectattr0_.ID as ID1116_, objectattr0_.AttributeValue as Attribut2_1116_, objectattr0_.AttributeID as Attribut3_1116_, objectattr0_1_.AttributeName as Attribut2_1117_ from ObjectAttributes objectattr0_ inner join Attributes objectattr0_1_ on objectattr0_.ID=objectattr0_1_.AttributeID 
10/14 16:36:39 [jrpp-12] HIBERNATE ERROR - [Macromedia] [SQLServer JDBC Driver][SQLServer]Invalid column name 'AttributeID'. 
10/14 16:36:39 [jrpp-12] HIBERNATE ERROR - [Macromedia] [SQLServer JDBC Driver][SQLServer]Statement(s) could not be prepared.

以下是查询的违规部分：

from ObjectAttributes objectattr0_ 
inner join Attributes objectattr0_1_ on objectattr0_.ID=objectattr0_1_.AttributeID

应该是：

from ObjectAttributes objectattr0_ 
inner join Attributes objectattr0_1_ on objectattr0_.AttributeID=objectattr0_1_.ID

ObjectAttribute.cfc上的AttributeName属性是导致问题的属性：

component  output="false" persistent="true" table="ObjectAttributes" 
{ 
        property name="ID" column="ID" generator="native" type="numeric" ormtype="int" fieldtype="id" unsavedvalue="0" ; 
        property name="AttributeValue" type="string" ; 
        property name="Attribute" fieldtype="many-to-one" cfc="Attribute" fkcolumn="AttributeID" fetch="join"; 
        property name="AttributeName" table="Attributes" joincolumn="AttributeID" ; 
}

我也尝试使用公式来获取ObjectAttribute实体上的AttributeName，如下所示：

component  output="false" persistent="true" table="ObjectAttributes"
{
    property name="ID" column="ID" generator="native" type="numeric" ormtype="int" fieldtype="id" unsavedvalue="0" ;
    property name="AttributeValue" type="string" ;
    property name="Attribute" fieldtype="many-to-one" cfc="Attribute" fkcolumn="AttributeID" fetch="join";
    property name="AttributeName" type="string" formula="(SELECT A.AttributeName FROM Attributes A WHERE A.ID = AttributeID)";
}

这有效，但我不能按计算列排序。如果我然后像这样调整PartObject.cfc：

property name="Attributes" cfc="ObjectAttribute" type="array" fkcolumn="ObjectID" fieldtype="one-to-many" orderby="AttributeName";

我在hibernatesql日志中遇到以下错误：

10/17 16:51:55 [jrpp-0] HIBERNATE DEBUG - select attributes0_.ObjectID as ObjectID2_, attributes0_.ID as ID2_, attributes0_.ID as ID244_1_, attributes0_.AttributeValue as Attribut2_244_1_, attributes0_.AttributeID as Attribut3_244_1_, ((SELECT A.AttributeName FROM Attributes A WHERE A.ID = attributes0_.AttributeID)) as formula25_1_, attribute1_.ID as ID246_0_, attribute1_.AttributeName as Attribut2_246_0_ from ObjectAttributes attributes0_ left outer join Attributes attribute1_ on attributes0_.AttributeID=attribute1_.ID where attributes0_.ObjectID=? order by attributes0_.AttributeName
10/17 16:51:55 [jrpp-0] HIBERNATE ERROR - [Macromedia][SQLServer JDBC Driver][SQLServer]Invalid column name 'AttributeName'.
10/17 16:51:55 [jrpp-0] HIBERNATE ERROR - [Macromedia][SQLServer JDBC Driver][SQLServer]Statement(s) could not be prepared.

这是一个不带的转储，该属性用于显示其余关系是否正常工作：

http://dl.dropbox.com/u/2586403/ORMIssues/Dump.pdf

我不知道如何解决这个问题。您将提供的任何帮助将不胜感激。

谢谢，

丹

Answer 1

我在Sam的帮助下解决了这个问题，在我的服务中设置了一个按照我想要的顺序返回项目的方法。我只是使用ORMExecuteQuery以正确的顺序获取项目，如果没有项目，只返回一个空数组。

最终方法看起来像这样，其中的规范直接按照我想要的顺序设置在对象中：

/**
@hint Gets a SpecGroups object based on ID. Pass 0 to retrieve a new empty SpecGroups object
@ID the numeric ID of the SpecGroups to return
@roles Admin, User
*/
remote ORM.SpecGroups function getSpecGroup(required numeric ID){
    if(Arguments.ID EQ 0){
        return New ORM.SpecGroups();
    }else{
        LOCAL.SpecGroup = EntityLoadByPK("SpecGroups", Arguments.ID);
        LOCAL.SpecsInGroup = ORMExecuteQuery("SELECT Spec FROM SpecInGroup G WHERE G.SpecGroupID = :GroupID ORDER BY SpecLabel, SpecName", {GroupID = LOCAL.SpecGroup.getID()});
        LOCAL.SpecGroup.setSpecifications(LOCAL.SpecsInGroup);
        return LOCAL.SpecGroup;
    }
}

使用ColdFusion ORM对类型表中的列进行排序

1 个答案: