Convert Hibernate @Formula to JOOQ field

问题

I am rewriting entire DB access layer from Hibernate to JOOQ and I face following issue.

One of JPA models is annotated with @Formula annotation as follows:

@Formula("fee1 + fee2 + fee3 + fee4")
private BigDecimal fee5;

Later in the code, a JPA query is made against the database which compares fee5 to parameter:

SELECT ... FROM ... WHERE fee5 > input;

How can above query be translated to JOOQ DSL?

回答1:

I managed to resolve the issue with following JOOQ query:

BigDecimal input = ...;
Field<BigDecimal> fee5 = TABLE.FEE1.add(TABLE.FEE2).add(TABLE.FEE3).add(TABLE.FEE4).as("fee5");
Condition cond = fee5.greaterThan(input);

来源：https://stackoverflow.com/questions/28166107/convert-hibernate-formula-to-jooq-field