Play框架获取数据单个字段与获取部分字段集合

Play 框架不提供针对某一个字段求和方法，那么一般出现这种情况都会去写原生sql去执行，非常简单：

String sql = "select sum(count) as count from book ";
Long sum;
Object sumFlag = JPA.em().createNativeQuery(sql).getSingleResult();

嗯，就这样，此方法同样适合其他返回一个字段的操作

jpa获取集合(如果你想只要自己想要的字段，那么可以如下获取)：
第一种：

	Query query = JPA.em().createNativeQuery("select id,name,price from book");
	//此段必加，指定返回的类型为Map
	query.unwrap(SQLQuery.class).setResultTransformer(Transformers.ALIAS_TO_ENTITY_MAP);
	List list = query.getResultList();
	for (int i = 0; i < list.size(); i++) {
		Map<String, Object> map = (Map<String, Object>) list.get(i);
		System.out.println(map.get("id"));
		System.out.println(map.get("name"));
		System.out.println(map.get("price"));
	}

第二种：

Query query = CurriVideo.em().createNativeQuery
        				("select "
        				+ "id               ,"
        				+ "name             ,"
        				+ "cover            ,"
        				+ "file_path        ,"
        				+ "charge           ,"
        				+ "price_money      ,"
        				+ "price_inte       ,"
        				+ "old_price_money  ,"
        				+ "old_price_inte   ,"
        				+ "null as publish_date     ,"
        				+ "null as status           ,"
        				+ "null as curri_id         ,"
        				+ "null as sort_num         "
        				+ "from curri_video",CurriVideo.class);
List<CurriVideo> list = query.getResultList();
for (CurriVideo curriVideo : list) {
	System.out.println(curriVideo.name);
	System.out.println(curriVideo.sort_num);
}

看实际业务来决定用哪种。
个人建议第一种，第二种有局限性，比如实体中有一字段为

@OneToMany
@JoinColumn(name="curri_id")
public List<Curriculum> currList;

那么必须查询出curri_id字段，并且不能设置为空，违背了我们最初想要的结果。