try {
  Query q = em.createQuery("SELECT t FROM Teacher AS t WHERE t.name like:name");
    q.setParameter("name", "Am%");
    System.out.println("Size:"+q.getResultList().size());
  } catch (Exception e) {
    e.printStackTrace();
  }

java.lang.IllegalArgumentException: An exception occurred while creating a query in EntityManager: 
Exception Description: Syntax error parsing [SELECT t FROM Teacher AS t WHERE t.name like:name]. 
[33, 49] The expression is not a valid conditional expression.
	at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1512)
	at test.Test.main(Test.java:38)
Caused by: Exception [EclipseLink-0] (Eclipse Persistence Services - 2.4.0.v20120608-r11652): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Syntax error parsing [SELECT t FROM Teacher AS t WHERE t.name like:name]. 
[33, 49] The expression is not a valid conditional expression.
	at org.eclipse.persistence.internal.jpa.jpql.HermesParser.buildException(HermesParser.java:150)
	at org.eclipse.persistence.internal.jpa.jpql.HermesParser.validate(HermesParser.java:325)
	at org.eclipse.persistence.internal.jpa.jpql.HermesParser.populateQueryImp(HermesParser.java:270)
	at org.eclipse.persistence.internal.jpa.jpql.HermesParser.buildQuery(HermesParser.java:157)
	at org.eclipse.persistence.internal.jpa.EJBQueryImpl.buildEJBQLDatabaseQuery(EJBQueryImpl.java:138)
	at org.eclipse.persistence.internal.jpa.EJBQueryImpl.buildEJBQLDatabaseQuery(EJBQueryImpl.java:112)
	at org.eclipse.persistence.internal.jpa.EJBQueryImpl.<init>(EJBQueryImpl.java:98)
	at org.eclipse.persistence.internal.jpa.EJBQueryImpl.<init>(EJBQueryImpl.java:82)
	at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1510)
	... 1 more