java - JPA:QueryCriteria where 子句中的谓词和表达式

Question

我有一种情况，在我的 where 子句中只有一个谓词和表达式。两者都需要在 where 子句中进行 ANDed:

Expression<String> col1 = tableEntity.get("col1");
Expression<String> regExpr = criteriaBuilder.literal("\\.\\d+$");
Expression<Boolean> regExprLike = criteriaBuilder.function("regexp_like", Boolean.class, col, regExpr);

Expression<TableEntity> col2= tableEntity.get("col2");
Predicate predicateNull = criteriaBuilder.isNull(col2);

createQuery.where(cb.and(predicateNull));
createQuery.where(regExprLike);

在这种情况下，我无法执行以下操作:createQuery.where(predicateNull, regExprLike);

我尝试使用 CriteriaBuilder 的 isTrue() 方法:

Predicate predicateNull = criteriaBuilder.isNull(col2);
Predicate predicateTrue = criteriaBuilder.isTrue(regExprLike);
createQuery.where(predicateNull, predicateTrue);

但这并没有帮助。

CriteriaQuery 允许在 where 子句中使用谓词或表达式，但不能同时使用两者。知道如何在 QueryCriteria 的 where 子句中同时使用谓词和表达式吗？

2014 年 10 月 10 日更新:按照 Chris 的建议，我尝试使用:

createQuery.where(predicateNull, regExprLike);

但我的查询失败并出现异常:

Caused by: org.jboss.arquillian.test.spi.ArquillianProxyException: org.hibernate.hql.internal.ast.QuerySyntaxException : unexpected AST node: ( near line 1, column 311 [select coalesce(substring(generatedAlias0.col1,0,(locate(regexp_substr(generatedAlias0.col1, :param0),
generatedAlias0.col1)-1)), generatedAlias0.col1), generatedAlias0.col1 
from com.temp.TableEntity as generatedAlias0 
where (generatedAlias0.col2 is null ) and ( regexp_like(generatedAlias0.col1, :param1))] [Proxied because : Original exception not deserilizable, ClassNotFoundException]

我的代码如下:

CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
CriteriaQuery<Object[]> createQuery = criteriaBuilder.createQuery(Object[].class);

Root<TableEntity> tableEntity = createQuery.from(TableEntity.class);

Expression<String> path = tableEntity.get("col1");

Expression<String> regExpr = criteriaBuilder.literal("\\.\\d+$");
Expression<String> regExprSubStr = criteriaBuilder.function("regexp_substr", String.class, path, regExpr);

Expression<Boolean> regExprLike = criteriaBuilder.function("regexp_like", Boolean.class, path, regExpr);


Expression<Integer> l3 = criteriaBuilder.locate(path, regExprSubStr);
Expression<Integer> minusOne = criteriaBuilder.literal(1);
Expression<Integer> l3Sub1 = criteriaBuilder.diff(l3, minusOne);
Expression<Integer> zeroIndex = criteriaBuilder.literal(0);
Expression<String> s3 = criteriaBuilder.substring(path, zeroIndex, l3Sub1);

Expression<TableEntity> col1 = tableEntity.get("col1");
Expression<TableEntity> col2 = tableEntity.get("col2");

Expression<String> coalesceExpr = criteriaBuilder.coalesce(s3, path);
createQuery.multiselect(coalesceExpr, col1);

Predicate predicateNull = criteriaBuilder.isNull(col2);

createQuery.where(criteriaBuilder.and(predicateNull, regExprLike));
String query = entityManager.createQuery(createQuery).unwrap(org.hibernate.Query.class).getQueryString();

Answer 1

我认为您的问题是 oracle 没有将“regexp_like”归类为函数。要使其工作，您必须使用新的注册函数扩展 Oracle 方言:

 public class Oracle12cExtendedDialect extends Oracle12cDialect {

public Oracle12cExtendedDialect() {
    super();
    registerFunction(
            "regexp_like", new SQLFunctionTemplate(StandardBasicTypes.BOOLEAN,
                    "(case when (regexp_like(?1, ?2)) then 1 else 0 end)")
    );
}
}

然后您可以更改您的 where 子句:

        createQuery.where(criteriaBuilder.and(predicateNull, criteriaBuilder.equal(regExprLike, 1)));

当然，你必须在 persistence.xml 中注册你的新方言

            <property name="hibernate.dialect" value="path.to.your.dialect.class.Oracle12cExtendedDialect" />