Question

    EntityManagerFactory entityManagerFactory = Persistence
            .createEntityManagerFactory("PrimeEclipseLink");
    EntityManager entityManager = entityManagerFactory
            .createEntityManager();
    Query query=entityManager.createQuery("select name from member");
    List<String> memberList=query.getResultList();
    for(String e:memberList){
        System.out.println(e);

我想在我的数据库中列出名称，但是我收到这样的错误消息。

Tem 14, 2015 9:55:28 AM com.sun.faces.lifecycle.InvokeApplicationPhase  execute
WARNING: #{memberControl.login()}: java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager: 
Exception Description: Syntax error parsing [select name from member]. 
[23, 23] An identification variable must be provided for a range variable declaration.
javax.faces.FacesException: #{memberControl.login()}:  java.lang.IllegalArgumentException: An exception occurred while creating a query    in EntityManager: 
Exception Description: Syntax error parsing [select name from member]. 
[23, 23] An identification variable must be provided for a range variable declaration.

Answer 1

尝试在以下内容中使用createNativeQuery代替createQuery：

Query query=entityManager.createNativeQuery("select name from member");

Answer 2

消息显示您的JPQL查询无效。只需将您的JPQL查询更改为

即可

SELECT m.name FROM member m

然后它会工作（因为这是有效的JPQL）。另一个答案意味着你抛弃数据存储独立性。

Jpa语法错误解析

2 个答案: