java.lang.IllegalArgumentException：名称为[xxx]的参数不存在
我正在使用Spring-data-jpa + hibernate
String login = "AdanaKebap";
String userQuery = "select * from user where username like '%:login%'"; // not working
Query query = entityManager.createNativeQuery(userQuery);
query.setParameter("login", login);
List<Object[]> userObjects = query.getResultList();
我会尝试
String userQuery = "select * from user where username like '%?%'"; // not working
query.setParameter("1", login);
String userQuery = "select * from user where username like '%?1%'"; // not working
query.setParameter("1", login);
String userQuery = "select * from user where username like ':login'"; // not working
query.setParameter("login", login);
String userQuery = "select * from user where username like :login "; // not working
query.setParameter("login", login);
String userQuery = "select * from user where username = :login "; // working, but i don't need this
最佳答案
您需要使用CONCAT来合并%和传递给查询的值，因此请尝试：
String userQuery = "select * from user where username like CONCAT('%', ?1, '%')";
//...
query.setParameter(1, login); // note here you have to use the index of placeholder
要么
String userQuery = "select * from user where username like CONCAT('%', :login, '%')'";
//...
query.setParameter("login", login);