JHipster Search entity by field of his entity

I have Tournament entity. He have OneToOne relation with Prize entity. Prize entity have filed "amount". So if i want to do search Tournaments that have prize between some 2 values how can i do that using JHipster QueryService ?

回答1:

Based on the fact you are using QueryService, I'm assuming you enabled the Filtering option when generating your entities. To query Tournament entities with a Prize amount between two values, a few things need to be added.

In TournamentQueryService, add the following to build a specification for the prize.amount field:

if (criteria.getPrizeAmount() != null) {
    specification = specification.and(buildReferringEntitySpecification(criteria.getPrizeAmount(), Tournament_.prize, Prize_.amount));
}

In TournamentCriteria, add a field for the prizeAmount filter: private DoubleFilter prizeAmount;. Also add getters and setters.

Now you can test the request through Swagger (under menu Admin->API) and filter tournaments based on the prize amount field.

If you want to make this query in the client, you need to add two parameters to the HTTP request:

prizeAmount.greaterOrEqualThan
prizeAmount.lessOrEqualThan

You can add them in the query call like below, which will return Tournaments with prize amounts >= 1 and <= 5 (example is Angular):

loadAll() {
    this.tournamentService.query({
        'prizeAmount.greaterOrEqualThan': 1,
        'prizeAmount.lessOrEqualThan': 5
    }).subscribe(
        (res: HttpResponse<ITournament[]>) => {
            this.tournaments = res.body;
        },
        (res: HttpErrorResponse) => this.onError(res.message)
    );
}

If you want the TournamentDTO to contain the prize amount as a field, the prizeAmount field needs to be added in the TournamentDTO (add field and getters/setters). You also need to add @Mapping(source = "prize.amount", target = "prizeAmount") in TournamentMapper to map the prize data to that field in the TournamentDTO.

回答2:

Is it possible to make that if we add another relation in Prize (Currency for example) and build specification ?

I didn't succeed because the definition is

buildReferringEntitySpecification(Filter<X> filter, SingularAttribute<? super ENTITY, OTHER> reference, SingularAttribute<OTHER, X> valueField)

So we have buildReferringEntitySpecification(Filter<Long> filter, SingularAttribute<Tournament, Prize> reference, SingularAttribute<Prize, Currency *(instead of Long)*> valueField)