JPA查詢LIKE不起作用

Question

此查詢有什么問題： SELECT i FROM Item i WHERE 1 = 1 AND UPPER(i.nome) LIKE %:nome% ORDER BY UPPER(i.nome) ？ 我嘗試在LIKE'％：nome％'中使用單引號和不使用單引號，但是它不起作用。

Exception in thread "JavaFX Application Thread" java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager: 
Exception Description: Syntax error parsing [SELECT i FROM Item i WHERE 1 = 1  AND UPPER(i.nome) LIKE %:nome%  ORDER BY UPPER(i.nome)]. 
[56, 63] The identification variable '%:nome%' is not following the rules for a Java identifier.
    at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1605)

Answer 1

正如解釋在這里 ，你需要包裝的價值% ，而設置參數，例如：

query = em.createQuery("SELECT i FROM Item i WHERE 1 = 1  AND UPPER(i.nome) LIKE :nome  ORDER BY UPPER(i.nome)");
query.setParameter("nome","%"+nome.toUpperCase()+"%");