小编Ali*_*lan的帖子

[dispatcherServlet] in context with path [] threw exception [Request processing failed: org.springframework.dao.InvalidDataAccessApiUsageException: Can't compare test expression of type [Facility] with element of type [basicType@4(java.lang.Integer,4)]] with root cause
java.lang.IllegalArgumentException: Can't compare test expression of type [Facility] with element of type [basicType@4(java.lang.Integer,4)]
    at org.hibernate.query.sqm.internal.SqmCriteriaNodeBuilder.assertComparable(SqmCriteriaNodeBuilder.java:2098)
    at org.hibernate.query.sqm.internal.SqmCriteriaNodeBuilder.equal(SqmCriteriaNodeBuilder.java:2121)
    at org.hibernate.query.sqm.internal.SqmCriteriaNodeBuilder.equal(SqmCriteriaNodeBuilder.java:182)
    at com.freit.medicalservice.repository.specification.PatientSpecification.lambda$facilityIdEqual$2a154287$1(PatientSpecification.java:139)
    at org.springframework.data.jpa.domain.SpecificationComposition.toPredicate(SpecificationComposition.java:62)
    at org.springframework.data.jpa.domain.SpecificationComposition.lambda$composed$e1dcbe07$1(SpecificationComposition.java:49)
    at org.springframework.data.jpa.domain.SpecificationComposition.toPredicate(SpecificationComposition.java:62)

在我从 spring 及其相关包（包括 JPA、Hibernate）从版本 5 迁移到版本 6 后，我在 JPA 规范上遇到了这个愚蠢的错误，这两天一直困扰着我。

我遇到错误的线路；

public static Specification<Patient> facilityIdEqual(Integer facilityId) {
        return (root, query, builder) -> {
            if (facilityId != …