Question

我正在创建一个在hibernate HQL中使用in indices函数的示例，方法是hibernate documentation

该文档的HQL示例如下：

from Show show where 'fizard' in indices(show.acts)

所以我为Show创建了一个实体，其中包含acts列表，如下所示：

@Entity
public class Show {
    @Id
    @GeneratedValue
    private int id;

    private String name;

    @ElementCollection
    private List<String> acts;

// Getters & Settters
}

当我尝试执行文档中提到的HQL时，我得到了例外：

Exception in thread "main" org.hibernate.QueryException: unindexed collection before indices() [from Show show where 'fizard' in indices(show.acts)]

所以为了使集合成为索引，我更新了我的Show实体属性acts，如下所示：

@ElementCollection
@OrderColumn
private List<String> acts;

在此之后我的程序开始抛出错误：

Hibernate: select show0_.id as id1_2_, show0_.name as name2_2_ from Show show0_ where 'fizard' in (select acts1_.acts_ORDER from Show_acts acts1_ where show0_.id=acts1_.Show_id)

ERROR: ORA-01722: invalid number

Oct 23, 2014 10:12:31 AM org.hibernate.engine.jdbc.connections.internal.DriverManagerConnectionProviderImpl stop
INFO: HHH000030: Cleaning up connection pool [jdbc:oracle:thin:@localhost:1521:xe]
Exception in thread "main" org.hibernate.exception.SQLGrammarException: could not extract ResultSet
    at org.hibernate.exception.internal.SQLExceptionTypeDelegate.convert(SQLExceptionTypeDelegate.java:80)
    at org.hibernate.exception.internal.StandardSQLExceptionConverter.convert(StandardSQLExceptionConverter.java:49)
    at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:126)
    at org.hibernate.engine.jdbc.spi.SqlExceptionHelper.convert(SqlExceptionHelper.java:112)
    at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.extract(ResultSetReturnImpl.java:91)
    at org.hibernate.loader.Loader.getResultSet(Loader.java:2065)
    at org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1862)
    at org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1838)
    at org.hibernate.loader.Loader.doQuery(Loader.java:909)
    at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:354)
    at org.hibernate.loader.Loader.doList(Loader.java:2553)
    at org.hibernate.loader.Loader.doList(Loader.java:2539)
    at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2369)
    at org.hibernate.loader.Loader.list(Loader.java:2364)
    at org.hibernate.loader.hql.QueryLoader.list(QueryLoader.java:496)
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.list(QueryTranslatorImpl.java:387)
    at org.hibernate.engine.query.spi.HQLQueryPlan.performList(HQLQueryPlan.java:231)
    at org.hibernate.internal.SessionImpl.list(SessionImpl.java:1264)
    at org.hibernate.internal.QueryImpl.list(QueryImpl.java:103)
    at AppTest.showData(AppTest3.java:44)
    at AppTest.main(AppTest3.java:29)
Caused by: java.sql.SQLSyntaxErrorException: ORA-01722: invalid number

    at oracle.jdbc.driver.T4CTTIoer.processError(T4CTTIoer.java:445)
    at oracle.jdbc.driver.T4CTTIoer.processError(T4CTTIoer.java:396)
    at oracle.jdbc.driver.T4C8Oall.processError(T4C8Oall.java:879)
    at oracle.jdbc.driver.T4CTTIfun.receive(T4CTTIfun.java:450)
    at oracle.jdbc.driver.T4CTTIfun.doRPC(T4CTTIfun.java:192)
    at oracle.jdbc.driver.T4C8Oall.doOALL(T4C8Oall.java:531)
    at oracle.jdbc.driver.T4CPreparedStatement.doOall8(T4CPreparedStatement.java:207)
    at oracle.jdbc.driver.T4CPreparedStatement.executeForDescribe(T4CPreparedStatement.java:884)
    at oracle.jdbc.driver.OracleStatement.executeMaybeDescribe(OracleStatement.java:1167)
    at oracle.jdbc.driver.OracleStatement.doExecuteWithTimeout(OracleStatement.java:1289)
    at oracle.jdbc.driver.OraclePreparedStatement.executeInternal(OraclePreparedStatement.java:3584)
    at oracle.jdbc.driver.OraclePreparedStatement.executeQuery(OraclePreparedStatement.java:3628)
    at oracle.jdbc.driver.OraclePreparedStatementWrapper.executeQuery(OraclePreparedStatementWrapper.java:1493)
    at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.extract(ResultSetReturnImpl.java:82)
    ... 16 more

以下是我正在处理的程序：

要将数据保存到DB中，我有以下逻辑：

private static void saveData() {
        Session session = getSession();
        session.getTransaction().begin();

        List<String> acts = new ArrayList<String>();
        acts.add("fizard");
        acts.add("Lord of Rings");

        Show s = new Show();
        s.setName("One");
        s.setActs(acts);

        List<String> acts1 = new ArrayList<String>();
        acts1.add("Tales");
        acts1.add("Stories");


        Show s1 = new Show();
        s1.setName("Two");
        s1.setActs(acts1);

        session.save(s);
        session.save(s1);

        session.getTransaction().commit();
        session.close();
    }

然后运行HQL我有以下逻辑：

private static void showData() {
        Session session = getSession();
        session.getTransaction().begin();

        Query q = session
                .createQuery("from Show show where 'fizard' in indices(show.acts)");
        List<Show> result = q.list();
        System.out.println("result=" + result);
        for (Show p : result) {
            System.out.println(p.getName());
        }

        session.getTransaction().commit();
        session.close();
    }

基于错误，我理解我正在尝试比较字符串'fizard'和索引是一个数字。但是，由于我只是关注官方的hibernate文档，有人可以告诉我如何使用hibernate的in indices函数编写一个简单的程序。

Answer 1

你质疑这个：休眠：从显示show0_选择show0_.id为id1_2_，show0_.name为name2_2_ 在哪里＆＃39; fizard＆＃39;在（从Show_acts acts1_中选择acts1_.acts_ORDER，其中show0_.id = acts1_.Show_id）;

感觉是可涂抹的，但acts1_.acts_ORDER可能是一个数字（例如数字）列，以获得这样的结果。

在indices（）函数中给出'ORA-01722：无效数'错误

1 个答案: