Re: JavaEOGenerator running, but I do not understand the output.
Re: JavaEOGenerator running, but I do not understand the output.
- Subject: Re: JavaEOGenerator running, but I do not understand the output.
- From: Chuck Hill <email@hidden>
- Date: Sun, 11 Nov 2007 10:03:17 -0800
On Nov 11, 2007, at 8:40 AM, Mike Schrag wrote:
After I made all the classes part of a package, everything was
fine, apart from the fact that they were 5.4-ish, and I want 5.3.
So I went to the second EOGenerator solution,
You can use JavaEOGenerator to produce 5.3 classes -- you just need
to modify the templates to not use 5.4 features.
I tried that, but there are some strange things: if there is no
class definition, eg it is just EOGenericRecord, it will create a
class EOGenericRecord, and _EOGenericRecord, or
com.webobjects.eocontrol.EOGenericRecord, depending on how the
thing is defined in the EOModel.
I don't ever use EOGenericRecord classes,
"I never use" :-)
And huh? You never what? You generate classes for those things?
but I thought that if you wanted this, that you should leave the
class names blank in the entity definition.
You need to specify it as EOGenericRecord for EOF.
Chuck
Also, it seems that there are some strange things going on with
n:m relations.
For instance, I have an n:m relation between product and review,
so there is a connection table reviewproduct, with a class
RevIewProduct.
Now I get an error in Product in this statement:
EOQualifier inverseQualifier = new EOKeyValueQualifier
(nl.immix.ReviewProduct.PRODUCT_KEY,
EOQualifier.QualifierOperatorEqual, this);
with the error:
nl.immix.ReviewProduct.PRODUCT_KEY cannot be resolved
which i scompletely logical as there is nno PRODUCT_KEY in the
ReviewProduct class.
Not sure I follow ... If you are expecting to be able to access
ReviewProduct.product(), you will need to make a class relationship
named "product" on your join class. You can't qualify on a non-
class relationship. If it was a class relationship, then the
templates would produce a PRODUCT_KEY for you.
ms
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