Getting column names from a JPA Native Query

Question

I have an administrative console in my web application that allows an admin to perform a custom SQL SELECT query on our database.

Underneath, the application is usin

Answer 1

Ryiad's answer DTO adds some confusion, you should have kept it away. You should have explained htat it works only with hibernate.

If like me you needs to keep the order of columns, you can specify your own transformer. i copied the code from hibernate and changed the HashMap to LinkedHashMap:

import java.util.LinkedHashMap;
import java.util.Map;

import org.hibernate.transform.AliasedTupleSubsetResultTransformer;
import org.hibernate.transform.ResultTransformer;

/**
 * {@link ResultTransformer} implementation which builds a map for each "row", made up of each aliased value where the
 * alias is the map key. Inspired by {@link org.hibernate.transform.AliasToEntityMapResultTransformer}, but kepping the
 * ordering of elements.
 * <p/>
 * Since this transformer is stateless, all instances would be considered equal. So for optimization purposes we limit
 * it to a single, singleton {@link #INSTANCE instance}.
 */
public class AliasToEntityMapResultTransformer extends AliasedTupleSubsetResultTransformer {

    public static final AliasToEntityMapResultTransformer INSTANCE = new AliasToEntityMapResultTransformer();

    /**
     * Disallow instantiation of AliasToEntityMapResultTransformer.
     */
    private AliasToEntityMapResultTransformer() {
    }

    @Override
    public Object transformTuple(Object[] tuple, String[] aliases) {
        Map result = new LinkedHashMap<>(tuple.length);
        for (int i = 0; i < tuple.length; i++) {
            String alias = aliases[i];
            if (alias != null) {
                result.put(alias, tuple[i]);
            }
        }
        return result;
    }

    @Override
    public boolean isTransformedValueATupleElement(String[] aliases, int tupleLength) {
        return false;
    }

    /**
     * Serialization hook for ensuring singleton uniqueing.
     *
     * @return The singleton instance : {@link #INSTANCE}
     */
    private Object readResolve() {
        return INSTANCE;
    }
}

With this transformer you can used Ryiad's solution with Hibernate:

    Query jpaQuery =  entityManager.createNativeQuery(queryString);
    org.hibernate.Query hibernateQuery =((org.hibernate.jpa.HibernateQuery)jpaQuery).getHibernateQuery();
  hibernateQuery.setResultTransformer(AliasToEntityMapResultTransformer.INSTANCE);
    List<Map<String,Object>> res = hibernateQuery.list();

Answer 2

2020

With hibernate 5.2.11.Final is actually pretty easy. In my example you can see how I get the column names for every row. And how I get values by column name.

Query q = em.createNativeQuery("SELECT columnA, columnB FROM table");
List<Tuple> result = q.getResultList();

for (Tuple row: result){

    // Get Column Names
    List<TupleElement<Object>> elements = row.getElements();
    for (TupleElement<Object> element : elements ) {
        System.out.println(element.getAlias());
    }

    // Get Objects by Column Name
    Object columnA;
    Object columnB;
    try {
        columnA = row.get("columnA");
        columnB= row.get("columnB");
    } catch (IllegalArgumentException e) {
        System.out.println("A column was not found");
    }
}

Answer 3

If the JPA provider does not support the retrieval of query metadata, another solution could be the use of a SQL parser like JSQLParser, ZQL or General SQL Parser (comercial), which extracts the fields from the SELECT statement.

Answer 4

This code worked for me

DTO Class :

 public class ItemResponse<T> {

 private T item;

 public ItemResponse() {
 }

 public ItemResponse(T item) {
   super();
   this.item = item;
 }

 public T getItem() {
    return item;
}

public void setItem(T item) {
    this.item = item;
}

}

Service Class is in the below

import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
import javax.persistence.Query;
import org.springframework.stereotype.Service;
import org.hibernate.transform.AliasToEntityMapResultTransformer;

@Service
public class ServiceClass{ 

@PersistenceContext
public EntityManager entityManager;

public ItemResponse exceuteQueryResponse(String queryString) {

        ItemResponse itemResponse=new ItemResponse();           
        Query jpaQuery =  entityManager.createNativeQuery(queryString);
        org.hibernate.Query hibernateQuery =((org.hibernate.jpa.HibernateQuery)jpaQuery).getHibernateQuery();
      hibernateQuery.setResultTransformer(AliasToEntityMapResultTransformer.INSTANCE);
        List<Map<String,Object>> res = hibernateQuery.list();

        itemResponse.setItem(res);
        return itemResponse;

    }

    }

Answer 5

cast query to hibernate query, then use hibernate method

          //normal use, javax.persistence.Query interface
    Query dbQuery = entityManager.createNativeQuery(sql);
    //cast to hibernate query
    org.hibernate.Query hibernateQuery =((org.hibernate.jpa.HibernateQuery)dbQuery)
            .getHibernateQuery();
    hibernateQuery.setResultTransformer(AliasToEntityMapResultTransformer.INSTANCE);

    List<Map<String,Object>> res = hibernateQuery.list();

    List<TxTestModel> txTestModels = new ArrayList<>();
    res.forEach(e->{
        TxTestModel txTestModel = new ObjectMapper().convertValue(e, TxTestModel.class);
    //  txTestModels.add(new TxTestModel().setIdd((Integer) e.get("idd")).setMmm((String) e.get("mmm")).setDdd((Date) e.get("ffffd")));
        txTestModels.add(txTestModel);
    });
    System.out.println(txTestModels.size());

Answer 6

I also faced a similar problem working with JPA. There is no direct way in JPA to access the resultset metadata. The solution can be extracting column names from the query itself or use JDBC to get the metadata.