本人想要实现的效果如下面sql 语句所示：

select * from user u where u.userid like '% 参数1 %' and u.username like '%参数2%'

通过百度已经实现 动态查询代码如下：

public Page getDataForSearch(PageRequest pageRequest, final JSONObject jSONObjSearch) throws SaggException {

Page page = null;

try {

page = userRepo.findAll(new Specification() {

@Override

public Predicate toPredicate(Root root, CriteriaQuery> query, CriteriaBuilder builder) {

List predicates = new ArrayList();

Predicate predicate = null;

Expression namePath = null;

if (null != jSONObjSearch) {

if (StrKit.notBlank(jSONObjSearch.getString("userid"))) {

namePath = root.get("userid");

predicate = builder.like(namePath, "%" + jSONObjSearch.getString("userid") + "%");

predicates.add(predicate);

}

if (StrKit.notBlank(jSONObjSearch.getString("userName"))) {

namePath = root.get("userName");

predicate = builder.like(namePath, "%" + jSONObjSearch.getString("userName") + "%");

predicates.add(predicate);

}

}

query.where(predicates.toArray(new Predicate[predicates.size()]));

return null;

}

}, pageRequest);

} catch (Exception e) {

// 交易人管理模块异常状态码statusCode用2000开始，2001：数据查询错误

throw new SaggException(StatusCodeUtil.personSearchErr);

}

return page;

}

DAO 层代码如下：

public interface UserRepository extends CrudRepository {

// List findAll(Pageable pageable);

Page findAll(Pageable pageable);// 分页需返回page对象

Page findAll(Specification spe, Pageable pageRequest);

Page findByUseridOrUserNameLike(Pageable pageable, String userid, String userName);

问题： 上面这种写法，好像不能达到我期望的效果，上面这个代码哪里错了？怎么改进？请大神赐教，感激不尽……