将 IN 子句列表添加到 JPA 查询

发布于 2024-10-06 10:53:21 字数 727 浏览 0 评论 0原文

我构建了一个如下所示的 NamedQuery:

@NamedQuery(name = "EventLog.viewDatesInclude",
        query = "SELECT el FROM EventLog el WHERE el.timeMark >= :dateFrom AND "
        + "el.timeMark <= :dateTo AND "
        + "el.name IN (:inclList)")

我想要做的是用项目列表而不是一个项目填充参数 :inclList。例如,如果我有一个 new List() { "a", "b", "c" } 如何在 :inclList 参数中获取它?它只让我编写一个字符串。例如:

setParameter("inclList", "a") // works

setParameter("inclList", "a, b") // does not work

setParameter("inclList", "'a', 'b'") // does not work

setParameter("inclList", list) // throws an exception

我知道我可以构建一个字符串并从中构建整个查询,但我想避免开销。有更好的方法吗?

相关问题:如果列表非常大,有什么好的方法来构建这样的查询吗?

I have built a NamedQuery that looks like this:

@NamedQuery(name = "EventLog.viewDatesInclude",
        query = "SELECT el FROM EventLog el WHERE el.timeMark >= :dateFrom AND "
        + "el.timeMark <= :dateTo AND "
        + "el.name IN (:inclList)")

What I want to do is fill in the parameter :inclList with a list of items instead of one item. For example if I have a new List<String>() { "a", "b", "c" } how do I get that in the :inclList parameter? It only lets me codify one string. For example:

setParameter("inclList", "a") // works

setParameter("inclList", "a, b") // does not work

setParameter("inclList", "'a', 'b'") // does not work

setParameter("inclList", list) // throws an exception

I know I could just build a string and build the whole Query from that, but I wanted to avoid the overhead. Is there a better way of doing this?

Related question: if the List is very large, is there any good way of building query like that?

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。

评论(4

傻比既视感 2024-10-13 10:53:21

IN 与集合值参数一起使用时,不需要 (...)

@NamedQuery(name = "EventLog.viewDatesInclude", 
    query = "SELECT el FROM EventLog el WHERE el.timeMark >= :dateFrom AND " 
    + "el.timeMark <= :dateTo AND " 
    + "el.name IN :inclList") 

When using IN with a collection-valued parameter you don't need (...):

@NamedQuery(name = "EventLog.viewDatesInclude", 
    query = "SELECT el FROM EventLog el WHERE el.timeMark >= :dateFrom AND " 
    + "el.timeMark <= :dateTo AND " 
    + "el.name IN :inclList") 
千里故人稀 2024-10-13 10:53:21

正确的 JPA 查询格式是:

el.name IN :inclList

如果您使用旧版本的 Hibernate 作为提供程序,则必须编写:

el.name IN (:inclList)

但这是一个错误 (HHH-5126)(编辑:现已解决)。

The proper JPA query format would be:

el.name IN :inclList

If you're using an older version of Hibernate as your provider you have to write:

el.name IN (:inclList)

but that is a bug (HHH-5126) (EDIT: which has been resolved by now).

北风几吹夏 2024-10-13 10:53:21
public List<DealInfo> getDealInfos(List<String> dealIds) {
        String queryStr = "SELECT NEW com.admin.entity.DealInfo(deal.url, deal.url, deal.url, deal.url, deal.price, deal.value) " + "FROM Deal AS deal where deal.id in :inclList";
        TypedQuery<DealInfo> query = em.createQuery(queryStr, DealInfo.class);
        query.setParameter("inclList", dealIds);
        return query.getResultList();
    }

适用于我的 JPA 2、Jboss 7.0.2

public List<DealInfo> getDealInfos(List<String> dealIds) {
        String queryStr = "SELECT NEW com.admin.entity.DealInfo(deal.url, deal.url, deal.url, deal.url, deal.price, deal.value) " + "FROM Deal AS deal where deal.id in :inclList";
        TypedQuery<DealInfo> query = em.createQuery(queryStr, DealInfo.class);
        query.setParameter("inclList", dealIds);
        return query.getResultList();
    }

Works for me with JPA 2, Jboss 7.0.2

ヤ经典坏疍 2024-10-13 10:53:21

您应该转换为 List ,如下所示:

    String[] valores = hierarquia.split(".");       
    List<String> lista =  Arrays.asList(valores);
    
    String jpqlQuery = "SELECT a " +
            "FROM AcessoScr a " +
            "WHERE a.scr IN :param ";
    
    Query query = getEntityManager().createQuery(jpqlQuery, AcessoScr.class);                   
    query.setParameter("param", lista);     
    List<AcessoScr> acessos = query.getResultList();

You should convert to List as shown below:

    String[] valores = hierarquia.split(".");       
    List<String> lista =  Arrays.asList(valores);
    
    String jpqlQuery = "SELECT a " +
            "FROM AcessoScr a " +
            "WHERE a.scr IN :param ";
    
    Query query = getEntityManager().createQuery(jpqlQuery, AcessoScr.class);                   
    query.setParameter("param", lista);     
    List<AcessoScr> acessos = query.getResultList();
~没有更多了~
我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
原文