Spring Data JPA规范以选择特定的列 [英] Spring Data JPA Specification to Select Specific Columns

问题描述

我们可以通过在存储库接口中编写自定义@Query方法来选择特定的列.但是，我不想为不同的属性编写这么多方法.

We can select specific columns by writing custom @Query methods in our Repository Interface. However, I don't want to write so many methods for different properties.

我尝试过此方法，但它始终会返回整个对象.

I tried this, but it returns the entire object all the time.

public class MySpecifications {

    public static Specification<MyInfo> propertiesWithId(final String[] properties, final Object id, final String idProperty)
    {

        return new Specification<MyInfo>() {

            @Override
            public Predicate toPredicate(Root<MyInfo> root,
                    CriteriaQuery<?> query, CriteriaBuilder cb) {

                query = cb.createTupleQuery(); //tried cb.createQuery(MyInfo.class); as well

                List<Selection<? extends Object>> selectionList = new ArrayList<Selection<? extends Object>>();

                for (String property : properties) {

                    Selection<? extends Object> selection = root.get(property);

                    selectionList.add(selection);
                }

                return query.multiselect(selectionList).where(cb.equal(root.get(idProperty), id)).getRestriction();
            }

        };
    }
}

用作:

MyInfo findOne(Specification(properties,idValue, idProperty));

这是正确的方法吗?错误在哪里?

Is this the correct way? Where is the mistake?

推荐答案

当前的spring data jpa规范执行器仅限于where子句中的条件，因此您不能更改所选的列，它隐式地仅限于完整实体(请参阅JpaSpecificationExecutor接口文档).您将必须使用自定义存储库实现，或转到命名查询-

The current spring data jpa specification executor is limited to criteria in the where clause, so you can't change the selected columns, it's implicitely limited to full entities only (take a look at JpaSpecificationExecutor interface documentation). You'll have to go with a custom repository implementation, or move to named queries-

春季数据JPA和Querydsl使用bean/构造函数投影来获取列的子集

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