Question

Finder def

@RooJpaActiveRecord(finders = {"findIndividualJobsByIndividualAndIndividualJobType"})

模型字段

@Enumerated(EnumType.STRING)
@ElementCollection(targetClass=IndividualJobType.class)
private List<IndividualJobType> individualJobType;

Roo正在生成这个Finder，它混合了String＆amp; StringBuffer的

    public static TypedQuery<IndividualJob> IndividualJob.findIndividualJobsByIndividualAndIndividualJobType(Individual individual, List<IndividualJobType> individualJobType, String sortFieldName, String sortOrder) {
    if (individual == null) throw new IllegalArgumentException("The individual argument is required");
    if (individualJobType == null) throw new IllegalArgumentException("The individualJobType argument is required");
    EntityManager em = IndividualJob.entityManager();
    StringBuilder queryBuilder = new StringBuilder("SELECT o FROM IndividualJob AS o WHERE o.individual = :individual AND");
    for (int i = 0; i < individualJobType.size(); i++) {
        if (i > 0) queryBuilder.append(" AND");
        queryBuilder.append(" :individualJobType_item").append(i).append(" MEMBER OF o.individualJobType");
    }
    if (fieldNames4OrderClauseFilter.contains(sortFieldName)) {
        jpaQuery = jpaQuery + " ORDER BY " + sortFieldName;
        if ("ASC".equalsIgnoreCase(sortOrder) || "DESC".equalsIgnoreCase(sortOrder)) {
            queryBuilder.append(" " + sortOrder);
        }
    }
    TypedQuery<IndividualJob> q = em.createQuery(queryBuilder.toString(), IndividualJob.class);
    q.setParameter("individual", individual);
    int individualJobTypeIndex = 0;
    for (IndividualJobType _individualjobtype: individualJobType) {
        q.setParameter("individualJobType_item" + individualJobTypeIndex++, _individualjobtype);
    }
    return q;
}

其他非Enum ElementCollections正确生成。

有什么想法吗？

Answer 1

我认为你发现了一个错误。你可以在https://jira.spring.io/browse/ROO/上报告一下吗？（谢谢！）

另一方面，解决方法可能是 push-in .java 中的方法并修复手工编写代码（这是Roo的一个重点，你不会被它生成的代码链接起来;-)）

祝你好运！

Spring Roo生成无效的查找程序？

1 个答案: