如何让 jOOQ 生成不引用主键的复合外键？

Question

Using the latest jOOQ with H2 1.4.194 no foreign keys in Keys.java are generated for the following (reduced) schema:

CREATE TABLE t (a INT, b INT, PRIMARY KEY (a));
CREATE TABLE u (a INT, b INT, FOREIGN KEY (a,b) REFERENCES t (a,b));

Adding a unique constraint as follows doesn't help:

CREATE TABLE t (a INT, b INT, PRIMARY KEY (a), UNIQUE (a,b));
CREATE TABLE u (a INT, b INT, FOREIGN KEY (a,b) REFERENCES t (a,b));

Changing the FK to reference the primary key (whether simple or composite) makes the FK appear.

Querying the FK via H2 works as expected.

Doesn't look like a regression (tried various 3.x jOOQ's), so wondering if I'm doing something wrong. Generator config (via maven plugin) as follows:

<name>org.jooq.util.JavaGenerator</name>
<database>
    <name>org.jooq.util.h2.H2Database</name>
    <includes>.*</includes>
    <excludes />
    <inputSchema>PUBLIC</inputSchema>
</database>

Answer 1

Working around missing foreign keys

Starting from jOOQ 3.14, if you're missing foreign keys from generated output either because of a limitation in jOOQ, or because you don't actually have any foreign keys in your schema, or because you're generating views, you can create synthetic objects , including synthetic primary keys , synthetic unique keys , and reference them from synthetic foreign keys with a configuration like this:

<configuration>
  <generator>
    <database>
      <syntheticObjects>
        <foreignKeys>
          <foreignKey>
            <tables>U</tables>
            <fields>
              <field>A</field>
              <field>B</field>
            </fields>
            <referencedTable>T</referencedTable>
            <referencedFields>
              <field>A</field>
              <field>B</field>
            </referencedFields>
          </foreignKey>
        </foreignKeys>
      </syntheticObjects>
    </database>
  </generator>
</configuration>

There are different ways to configure this, eg by name reference or by column reference, etc. check out the manual for details .

Historic answer (it was a bug)

The query that is run internally by jOOQ's code generator is this:

select 
  "CROSS_REFERENCES"."FK_NAME", 
  "CROSS_REFERENCES"."FKTABLE_NAME", 
  "CROSS_REFERENCES"."FKTABLE_SCHEMA", 
  "CROSS_REFERENCES"."FKCOLUMN_NAME", 
  "CONSTRAINTS"."CONSTRAINT_NAME", 
  "CONSTRAINTS"."CONSTRAINT_SCHEMA"
from "INFORMATION_SCHEMA"."CROSS_REFERENCES"
  join "INFORMATION_SCHEMA"."CONSTRAINTS"
  on (
    "CROSS_REFERENCES"."PK_NAME" = "CONSTRAINTS"."UNIQUE_INDEX_NAME"
    and "CROSS_REFERENCES"."PKTABLE_NAME" = "CONSTRAINTS"."TABLE_NAME"
    and "CROSS_REFERENCES"."PKTABLE_SCHEMA" = "CONSTRAINTS"."TABLE_SCHEMA"
  )
where (
  "CROSS_REFERENCES"."FKTABLE_SCHEMA" in (
    'PUBLIC'
  )
  and "CONSTRAINTS"."CONSTRAINT_TYPE" in (
    'PRIMARY KEY', 'UNIQUE'
  )
)
order by 
  "CROSS_REFERENCES"."FKTABLE_SCHEMA" asc, 
  "CROSS_REFERENCES"."FK_NAME" asc, 
  "CROSS_REFERENCES"."ORDINAL_POSITION" asc

The query appears correct, but there seems to be a misunderstanding in how H2 encodes unique constraints in these dictionary views. Or a bug in H2.

I've created two issues, let's see which one is right:

• Bug in H2: https://github.com/h2database/h2database/issues/466
• Bug in jOOQ: https://github.com/jOOQ/jOOQ/issues/5972