Question

我有一种情况，在我的where子句中我有单个谓词和表达式。两者都需要在where子句中进行AND运算：

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);

但它没有帮助。

CriteriaQuery在where子句中允许谓词或表达式，但不允许两者都允许。我怎么能在QueryCriteria的where子句中使用谓词和表达式？

2014年10月10日更新： 根据Chris的建议，我尝试使用：

createQuery.where(predicateNull, regExprLike);

但我的查询失败，例外：

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

我认为你的问题是oracle没有将'regexp_like'归类为函数。要使其工作，您必须使用新注册的函数扩展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)")
    );
}
}

然后你可以改变你的where子句：

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

当然，您必须在persistence.xml中注册新方言

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

JPA：谓词和表达式都在QueryCriteria where子句中

1 个答案: