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我有三个表,结构如下:

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

我正在处理的三个对象在这里:

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

我需要能够获取一个 PartObject,然后拉取它的所有属性,按 Types 表中的 AttributeName 排序。以下是我遇到的问题:

  1. 我无法通过其 Attribute.AttributeName 属性对 PartObject 中的 Attributes 属性进行排序

  2. 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休眠日志文件

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. 

这是查询的违规部分:

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

它应该是:

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

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

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" ; 
} 

我还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,但我无法按该计算列排序。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 日志中收到以下错误:

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.

这是一个没有该属性的转储,以表明其余关系正常工作:

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

我不知道如何解决这个问题。您能提供的任何帮助将不胜感激。

谢谢,

4

1 回答 1

0

我在 Sam 的帮助下解决了这个问题,方法是在我的 Service 中设置一个按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,则返回一个空数组。

最终方法看起来像这样,其中规范按照我想要的顺序直接在对象中设置:

/**
@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;
    }
}
于 2011-09-12T18:57:24.317 回答