Question

在PostgreSQL中， string_agg（列，分隔符）允许聚合一些字符串。我尝试将它与JPA一起使用，但它不是JPA标准函数。

注意：相当于 CriteriaBuilder＃concat（）。

所以，我试图告诉JPA这个函数存在，如下所示：

public class StringAgg extends ParameterizedFunctionExpression<String> implements Serializable {

  public static final String NAME = "string_agg";

  @Override
  public boolean isAggregation() {
    return true;
  }

  @Override
  protected boolean isStandardJpaFunction() {
    return false;
  }

  public StringAgg(CriteriaBuilderImpl criteriaBuilder, Expression<String> expression, String separator) {
    super(criteriaBuilder, String.class, NAME, expression, new LiteralExpression(criteriaBuilder, separator));
  }
}

然后：

Expression<String> exprStr = ...
CriteriaBuilder cb = ...
cb.construct(MyClass.class, 
             myClass.get(MyClass_.name),
             myClass.get(MyClass_.surname),
             new StringAgg(cb, exprStr, "/"));

问题，我收到NullPointerException ！

java.lang.NullPointerException: null
at org.hibernate.internal.util.ReflectHelper.getConstructor(ReflectHelper.java:355) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.hql.internal.ast.tree.ConstructorNode.resolveConstructor(ConstructorNode.java:179) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.hql.internal.ast.tree.ConstructorNode.prepare(ConstructorNode.java:152) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.hql.internal.ast.HqlSqlWalker.processConstructor(HqlSqlWalker.java:1028) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectExpr(HqlSqlBaseWalker.java:2279) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectExprList(HqlSqlBaseWalker.java:2145) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectClause(HqlSqlBaseWalker.java:1451) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.query(HqlSqlBaseWalker.java:571) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectStatement(HqlSqlBaseWalker.java:299) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.statement(HqlSqlBaseWalker.java:247) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.hql.internal.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:261) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:189) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:141) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:119) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:87) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.engine.query.spi.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:190) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.internal.AbstractSessionImpl.getHQLQueryPlan(AbstractSessionImpl.java:288) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]
at org.hibernate.internal.AbstractSessionImpl.createQuery(AbstractSessionImpl.java:223) ~[hibernate-core-4.3.0.Beta3.jar:4.3.0.Beta3]

调试程序显示Selection（cb.construct()）的最后new StringAgg(cb, exprStr, "/")被忽略。因此，搜索到的构造函数为MyClass(String,String)而不是MyClass(String, String, String)。

StringAgg的实现有什么问题吗？有人已经尝试在JPA中使用string_agg吗？

解决方案（感谢vzamanillo）

扩展方言：

public class PGDialect extends PostgreSQLDialect{

  public PGDialect() {
    super();
    registerFunction("string_agg", new SQLFunctionTemplate( StandardBasicTypes.STRING, "string_agg(?1, ?2)"));
  }
}

在 persistence.xml

中使用它

<properties>
  <property name="javax.persistence.jdbc.driver" value="org.postgresql.Driver"/>
  <property name="hibernate.dialect" value="path.to.PGDialect"/>

然后使用 CriteriaBuilder＃function（）：

Expression<String> exprStr = ...
CriteriaBuilder cb = ...
cb.construct(MyClass.class, 
             myClass.get(MyClass_.name),
             myClass.get(MyClass_.surname),
             cb.function( "string_agg", myColPath, cb.literal("delimiter" )));

为了简化它，我创建了一个辅助方法：

public static Expression<String> strAgg(CriteriaBuilder cb, Expression<String> expression, String delimiter) {
  return cb.function( "string_agg", String.class, expression, cb.literal(delimiter));
}

所以代码变成：

Expression<String> exprStr = ...
CriteriaBuilder cb = ...
cb.construct(MyClass.class, 
             myClass.get(MyClass_.name),
             myClass.get(MyClass_.surname),
             strAgg(cb, myColPath, "delimiter"));

Answer 1

也许这有助于你，

您可以在JPA Criteria Query中调用数据库函数。

CriteriaBuilder接口有一个“function”方法。

<T> Expression<T> function(String name,
                         Class<T> type,
                         Expression<?>... args)

Create an expression for the execution of a database function.

Parameters:
    name - function name
    type - expected result type
    args - function arguments
Returns:
    expression

然后你可以尝试创建一个CriteriaBuilder助手类来获得一个普通的标准表达式，你可以在我们的标准查询中照常使用

public abstract class CriteriaBuilderHelper {

    private static final String PG_STRING_AGG  = "string_agg";

    /**
    * @param cb the CriteriaBuilder to use
    * @param toJoin the string to join
    * @param delimiter the string to use
    * @return Expression<String>
    */
    public static Expression functionStringAgg(CriteriaBuilder cb, String toJoin, String delimiter) {
        return cb.function(PG_STRING_AGG, 
            String.class,
            cb.literal(toJoin),
            cb.literal(delimiter))
        );
    }
}

或者您可以使用自定义方言注册新功能

public class PGDialect extends PostgreSQLDialect{

    public PGDialect() {
        super();
        registerFunction("string_agg", new SQLFunctionTemplate( StandardBasicTypes.STRING, "string_agg(?1, ?2)"));
    }
}

并在CriteriaBuilder中将其用作普通函数

Expression<String> functionStringAgg = cb.function( "string_agg", String.class, 
                                cb.parameter(String.class, "toJoin" ), 
                                cb.parameter(String.class, "delimiter"));

毕竟不要忘记将参数值设置为CriteriaQuery

setParameter( "toJoin", toJoin);
setParameter( "delimiter", delimiter);

Answer 2

我使用相同的结构来完成它，但是很容易：

修改了持久性文件：

    <property name="jpaProperties">
        <props>
                <prop key="hibernate.dialect">es.gmrcanarias.saga.utiles.PGDialect</prop>
                <prop key="hibernate.show_sql">true</prop>
...
        </props>
    </property>

然后注册功能：

import org.hibernate.dialect.PostgreSQL82Dialect;
import org.hibernate.dialect.function.SQLFunctionTemplate;
import org.hibernate.type.StandardBasicTypes;

public class PGDialect extends PostgreSQL82Dialect {

   public PGDialect() {
       super();
      registerFunction("string_agg", new  
       SQLFunctionTemplate(StandardBasicTypes.STRING, "string_agg(?1, ?2)"));
   }
}

然后添加查询：

    Join<Razon, Incidenc> subquery;
    ....
    Expression<String> functionStringAgg = criteriaBuilder.function("string_agg", 
        String.class,
        subquery.get(CODIGO), 
        criteriaBuilder.literal(", "));
...
    subqueryList.select(functionStringAgg);

Answer 3

PGDialect的一个细节：您可以添加第3个参数以使用ORDER BY

public class PGDialect extends PostgreSQL9Dialect {


     public PGDialect() {
         super();
         this.registerFunction("string_agg", new SQLFunctionTemplate( StandardBasicTypes.STRING, "string_agg(?1, ?2)") );
         this.registerFunction("string_agg", new SQLFunctionTemplate( StandardBasicTypes.STRING, "string_agg(?1, ?2 ORDER BY ?3 )") );
     }
 }

HQL查询用法

 "SELECT string_agg(f.name, '; ', f.name) FROM Foo as f "

JPA中的PostgreSQL函数string_agg

3 个答案: