java - 基本路径使用异常 : Cannot join to attribute of basic type

Question

我正在使用 Java 8 和 JPA (Hibernate 5.2.1)。我有一个完美的 like 子句，直到我引入 like 子句以也使用外键表值。

我收到以下错误:

BasicPathUsageException: Cannot join to attribute of basic type

我认为问题与我有一个连接表有关，我不确定如何使用连接表创建 LIKE 连接。

员工 - employee_category - 类别

模型(Employee.java):

@ManyToMany(cascade = CascadeType.ALL, fetch=FetchType.EAGER)
@JoinTable
(
    name="employee_category",
    joinColumns={ @JoinColumn(name="EMP_ID", referencedColumnName="ID") },
    inverseJoinColumns={ @JoinColumn(name="CAT_ID", referencedColumnName="ID") }
)
private Set<Category> categories;

模型(Category.java):

@Id
private String id;

JPA

当我引入以下代码时，它失败并出现上述错误(我认为问题出在这段新代码上):

Join<T, Category> category = from.join("id");
predicates.add(criteriaBuilder.like(category.<String>get("name"), "%" + searchQuery + "%"));

这是整个方法:

protected List<T> findAllCriteria(String[] orderByAsc, String[] orderByDesc, Class<T> typeParameterClass,
        int firstResult, int maxResults, String searchQuery) {
    CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
    CriteriaQuery<T> criteriaQuery = criteriaBuilder.createQuery(typeParameterClass);

    // from
    Root<T> from = criteriaQuery.from(typeParameterClass);
    criteriaQuery.select(from);

    // like
    if (searchQuery != null && searchQuery.trim().length() > 0) {

        List<Predicate> predicates = new ArrayList<Predicate>();
        for (String name : getColumnNames(typeParameterClass)) {
            Predicate condition = criteriaBuilder.like(from.<String>get(name), "%" + searchQuery + "%");
            predicates.add(condition);
        }
        Join<T, Category> category = from.join("id");
        predicates.add(criteriaBuilder.like(category.<String>get("name"), "%" + searchQuery + "%"));

        criteriaQuery.where(criteriaBuilder.or(predicates.toArray(new Predicate[] {})));
    }

    List<T> results = (List<T>) entityManager.createQuery(criteriaQuery).setFirstResult(firstResult)
            .setMaxResults(maxResults).getResultList();
    return results;
}

我们将不胜感激。

Answer 1

已解决:加入类别 = from.join("categories");