Question

我有以下问题。我使用了spring-data，我想为Long Property创建一个“like”语句。

我尝试了以下内容，但没有成功。

private Specification<T> createSpecification()
{
    return new Specification<T>()
    {

        @Override
        public Predicate toPredicate(final Root<T> root, final CriteriaQuery<?> query, final CriteriaBuilder cb)
        {

            final Predicate like = query.where(cb.like(root.<String> get("kundenNr"), "%4%")).getRestriction();

            return like;
        }
    };

}

和域类

public class Kunde implements Serializable
{

@ColumnYB(lable = "KundenNr.", order = 1)
@Id
@javax.persistence.Column(nullable = false, name = "kundennr")
protected Long kundenNr;

@ColumnYB(lable = "Deaktiviert")
@Column(nullable = false)
private boolean deaktiviert = false;

@Column(name = "kundenart")
private String kundenart;
...

我收到以下错误。

org.springframework.dao.InvalidDataAccessApiUsageException: Parameter value [%4%] did not match expected type [java.lang.Long]; nested exception is java.lang.IllegalArgumentException: Parameter value [%4%] did not match expected type [java.lang.Long]
at org.springframework.orm.jpa.EntityManagerFactoryUtils.convertJpaAccessExceptionIfPossible(EntityManagerFactoryUtils.java:384) ~[spring-orm-4.0.1.RELEASE.jar:4.0.1.RELEASE]
at org.springframework.orm.jpa.vendor.HibernateJpaDialect.translateExceptionIfPossible(HibernateJpaDialect.java:157) ~[spring-orm-4.0.1.RELEASE.jar:4.0.1.RELEASE]
at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.translateExceptionIfPossible(AbstractEntityManagerFactoryBean.java:417) ~[spring-orm-4.0.1.RELEASE.jar:4.0.1.RELEASE]
at org.springframework.dao.support.ChainedPersistenceExceptionTranslator.translateExceptionIfPossible(ChainedPersistenceExceptionTranslator.java:59) ~[spring-tx-4.0.1.RELEASE.jar:4.0.1.RELEASE]
at org.springframework.dao.support.DataAccessUtils.translateIfNecessary(DataAccessUtils.java:213) ~[spring-tx-4.0.1.RELEASE.jar:4.0.1.RELEASE]
at org.springframework.dao.support.PersistenceExceptionTranslationInterceptor.invoke(PersistenceExceptionTranslationInterceptor.java:147) ~[spring-tx-4.0.1.RELEASE.jar:4.0.1.RELEASE]
at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:179) ~[spring-aop-4.0.1.RELEASE.jar:4.0.1.RELEASE]
at org.springframework.data.jpa.repository.support.LockModeRepositoryPostProcessor$LockModePopulatingMethodIntercceptor.invoke(LockModeRepositoryPostProcessor.java:92) ~[spring-data-jpa-1.4.1.RELEASE.jar:na]
at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:179) ~[spring-aop-4.0.1.RELEASE.jar:4.0.1.RELEASE]
at org.springframework.aop.interceptor.ExposeInvocationInterceptor.invoke(ExposeInvocationInterceptor.java:92) ~[spring-aop-4.0.1.RELEASE.jar:4.0.1.RELEASE]
at org.springframework.aop.framework.ReflectiveMethodInvocation.proceed(ReflectiveMethodInvocation.java:179) ~[spring-aop-4.0.1.RELEASE.jar:4.0.1.RELEASE]

我做错了什么？

Answer 1

这是因为您的实体类中的kundenNr是Long，而%4%的解析时间不长。

在SQL中，查询中的％符号仅适用于varchar（以及某些变体），因此无法在长字段上使用它，这可能会转换为数据库中的数字字段。

Answer 2

使用以下规范来查找其列（NUMBER）与某些数字匹配的记录：

    public static org.springframework.data.jpa.domain.Specification<Model> numberLikeValue(Long value) {
    return (root, query, builder) ->
            org.springframework.util.ObjectUtils.isEmpty(value)
                    ? builder.conjunction()
                    : builder.like(builder.function("TO_CHAR", String.class, root.get("dbColumn")), contains(String.valueOf(value)));
}

使用了Oracle数据库。

Spring Data JPA：带有“like”和Long的规范

2 个答案: