[英]JPA java.lang.IllegalArgumentException: Parameter with that name [xxx] did not exist
[英]java.lang.IllegalArgumentException: Parameter with that name [xxx] did not exist
java.lang.IllegalArgumentException:具有該名稱 [xxx] 的參數不存在
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();
我會嘗試
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
您需要使用CONCAT
來連接%
和傳遞給查詢的值,因此請嘗試:
String userQuery = "select * from user where username like CONCAT('%', ?1, '%')";
//...
query.setParameter(1, login); // note here you have to use the index of placeholder
或者
String userQuery = "select * from user where username like CONCAT('%', :login, '%')'";
//...
query.setParameter("login", login);
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