如何在JPA中创建对列表作为参数?

时间:2015-07-06 14:59:05

标签: java sql hibernate jpa

在我的Java应用程序中,我想创建SQL查询,最终将看起来像这样:

SELECT * FROM my_table t WHERE (t.a, t.b) IN ((a1,b1),(a2,b2),(a3,b3), ... )

如何生成?

我的代码如下所示:

public List<MyEntity> getMyEntity(List<Long> alist, List<Long> blist) {

    String stringQuery = "SELECT * FROM my_table t WHERE (t.a , t.b) in (:alist, :blist)"; 

    // This is kinda how I whould like my query to look, but I guess I will generate something like:
    // SELECT * FROM my_table t WHERE (t.a, t.b) IN ((a1, a2, a3, ...), (b1, b2, b3, ...))
    // which isn't the same and it isn't what I want 

    Query query = getEntityManager().createNativeQuery(stringQuery, MyEntity.class);

    // I'm using native query because in my aplication above query is in fact much more complicated, 
    // and I can't use standard JPA query. I cut those parts here, because they aren't directly related to my problem
    // in the other way that they force to use native query.

    query.setParameter("alist", alist);
    query.setParameter("blist", blist);

    return query.getResultList();
}

1 个答案:

答案 0 :(得分:1)

您使用本机查询,似乎本机查询根本没有列表参数:

Set list parameter to native query

How to use a dynamic parameter in a IN clause of a JPA named query?

您必须手动连接SQL。