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.
最佳答案
尝试在以下内容中使用createNativeQuery
而不是createQuery
:
查询 query=entityManager.createNativeQuery("从成员中选择名称");
关于java - Jpa语法错误解析,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31399775/