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Predicate throws IllegalArgumentException

See original GitHub issue

springboot version: 1.5.9 querydsl version: 4.1.4

service query code:

    public List<Timeline> searchAllByUserId(long userId, int pageIndex, int pageSize) {
        QTimeline qTimeline = QTimeline.timeline;
        Predicate predicate = qTimeline.userId.longValue().eq(userId);
        PageRequest page = new PageRequest(pageIndex, pageSize, new Sort(Sort.Direction.DESC, "createAt"));

        return timelineRepository.findAllByUserId(predicate, page);
    }

exception:

org.springframework.dao.InvalidDataAccessApiUsageException: Parameter value [timeline.userId = 2] did not match expected type [java.lang.Long (n/a)]; nested exception is java.lang.IllegalArgumentException: Parameter value [timeline.userId = 2] did not match expected type [java.lang.Long (n/a)]

Caused by: java.lang.IllegalArgumentException: Parameter value [timeline.userId = 2] did not match expected type [java.lang.Long (n/a)]
	at org.hibernate.jpa.spi.BaseQueryImpl.validateBinding(BaseQueryImpl.java:897)
	at org.hibernate.jpa.internal.QueryImpl.access$000(QueryImpl.java:61)
	at org.hibernate.jpa.internal.QueryImpl$ParameterRegistrationImpl.bindValue(QueryImpl.java:235)
	at org.hibernate.jpa.spi.BaseQueryImpl.setParameter(BaseQueryImpl.java:638)

anyone could give me some tips ?

Issue Analytics

State:
Created 5 years ago
Comments:8 (3 by maintainers)

Top GitHub Comments

1reaction

Shredder121commented, May 22, 2018

findAllByUserId

Spring data will generate a select ... where user.id = ? query. So you just pass in the id, not a predicate.

1reaction

linyongfu2013commented, May 22, 2018

so disappointed no one give a damn to new issue… @meatballhat @rocketraman @johnktims @ponzao @msiedlarek

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