Mocking non-Table fields?

[Deven Phillips]

I saw this:

https://groups.google.com/d/msg/jooq-user/C2LpNFUewdk/7RINa5ZDKBIJ

and the related issue:

https://github.com/jOOQ/jOOQ/issues/3139

But I don't think it addresses the question I have...

I am creating a query:

create.select(

        CUSTOMER.NAME.as("name"), 

        CUSTOMER.CODE.as("code"), 

        concat(val("/v3/customer/"), CUSTOMER.CODE, val("/vdc")).as("vdc"))

    .where(CUSTOMER.CODE.eq(code));

When I create a MockDataProvider, I can use CUSTOMER.CODE and CUSTOMER.NAME as field identifiers, but how do I identify the calculated field in the MockDataProvider?

E.g.:

Result<CustomerRecord> result = create.newResult(CUSTOMER);

result.add(create.newRecord(CUSTOMER));

result.get(0).setValue(CUSTOMER.NAME, "John Doe");

result.get(0).setValue(CUSTOMER.CODE, "29793408");

result.get(0).setValue(*****What can I put here?!?!*****, "/v3/customer/29793408/vdc");

Any help would be greatly appreciated!!

Deven

[Lukas Eder]

When you select "name", "code" and "vdc", you will also have to provide values for exactly the same fields, e.g. fieldByName("name"), fieldByName("code"), etc... Field identifiers are case-sensitive in jOOQ.

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[Deven Phillips]

Lukas,

    Yes, and in other iterations I have done that like:

result.get(0).setValue(CUSTOMER.NAME.as("name"), "John Doe");

But that still doesn't answer my real question of how to handle the calculated field? Or did I not understand you completely?

Thanks,

Deven

[Lukas Eder]

You have to provide a value for a field whose name is "vdc".

I understand the source of confusion now, though. You're creating a Result<CustomerRecord>, when you really want to create a Result<Record3<String, String, String>> for this very specific query.

Use DSLContext.newRecord(Field, Field, Field) to construct such records.

[Deven Phillips]

OK, I can follow that, but how do I create an instance of type "Field" with a name "vdc" and type of "String"? So far, I just see that I would have to create an anonymous implementation of the Field interface and implement all of the methods... Is there a better way?

Thanks in advance and thanks for the fast replies!!

Deven

[Deven Phillips]

Also, trying to do something like:

Result<Record3<Field<String>, Field<String>, Field<String>>> result = create.newResult(*****WHAT DO I PUT HERE*****);

That's problematic as well...

Deven

[Lukas Eder]

2014-06-16 20:31 GMT+02:00 Deven Phillips <deven.p...@gmail.com>:

OK, I can follow that, but how do I create an instance of type "Field" with a name "vdc" and type of "String"? So far, I just see that I would have to create an anonymous implementation of the Field interface and implement all of the methods... Is there a better way?

Thousands of ways :-)

For instance, DSL.fieldByName, as suggested previously:

http://www.jooq.org/javadoc/latest/org/jooq/impl/DSL.html#fieldByName(java.lang.String...)

 

Thanks in advance and thanks for the fast replies!!

No worries. I happen to do something quite boring next to answering mails just now ;-)

[Lukas Eder]

Indeed, the missing methods required to create such a Result will be added in jOOQ 3.4.0. You've already found the relevant issue for that:

https://github.com/jOOQ/jOOQ/issues/3139

You have at least these two options:

- Upgrading to a version checked out and built from GitHub

- Creating new org.jooq.impl.ResultImpl instances through reflection

[Deven Phillips]

Aha!!!! And enlightenment is found... Thank you!

You were EXCEPTIONALLY helpful.. I greatly appreciate the assistance!

Deven

[Lukas Eder]

Great, glad I could help you today,

Cheers

Lukas