例如,有2个属性门牌号和密码,我想要一个属性作为地址,如门牌号是10,pincode是110064和组合地址属性是10,110064这是我的代码
final Criteria criteria= getDatabaseSession().createCriteria(Application.class, "application");
final ProjectionList projectionList=Projections.projectionList();
criteria.setProjection(projectionList);
projectionList.add(Projections.property("address.street"), "street");
projectionList.add(Projections.property("address.postcode"), "postcode");
projectionList.add(Projections.property("address.houseNumber"), "houseNumber");
criteria.createAlias("application.applicationCase", "applicationCase", JoinType.INNER_JOIN);
criteria.createAlias("applicationCase.property", "property");
criteria.createAlias("property.address", "address");
criteria.setResultTransformer(Criteria.ALIAS_TO_ENTITY_MAP);
return (Map<String, Object>) criteria.uniqueResult();
我想做这样的事情
projectionList.add(Projections.property("address.street"+"address.houseNumber"+"address.postcode"),"address");
有人可以帮忙吗