1
我是Play新手，在產品模式下部署 應用程序時遇到問題。 我正在使用的內存數據庫，開始.... 玩的版本是1.2.4 我在控制器中做了一個簡單的應用程序與實體在產品模式下未發現播放幀工作表錯誤
@Entity
public class TestModel extends Model {
public String name;
}
，並作爲CONF
application.mode=prod
db=mem
%test.application.mode=prod
%test.db.url=jdbc:h2:mem:play;MODE=MYSQL;LOCK_MODE=0
%test.jpa.ddl=create
%test.mail.smtp=mock
只是爲了測試我在做什麼是
public static void listBars() {
TestModel bar = new TestModel();
bar.name = "a new bar";
bar.save();
TestModel bar2 = new TestModel();
bar2.name = "a new bar2";
bar2.save();
renderJSON(TestModel.findAll());
}
很感激，如果我能得到this..The代碼一些幫助在開發模式下工作..
的異常跟蹤是
va:116)
at org.hibernate.id.IdentityGenerator$GetGeneratedKeysDelegate.prepare(I
entityGenerator.java:90)
at org.hibernate.id.insert.AbstractReturningDelegate.performInsert(Abstr
ctReturningDelegate.java:54)
... 24 more
9:57:31,396 WARN ~ SQL Error: 42102, SQLState: 42S02
9:57:31,397 ERROR ~ Table "TESTMODEL" not found; SQL statement:
nsert into TestModel (id, name) values (null, ?) [42102-149]
9:57:31,404 ERROR ~
69772kodd
nternal Server Error (500) for request POST /bars.json
xecution exception (In /app/controllers/Application.java around line 30)
ersistenceException occured : org.hibernate.exception.SQLGrammarException: coul
not insert: [models.TestModel]
lay.exceptions.JavaExecutionException: org.hibernate.exception.SQLGrammarExcept
on: could not insert: [models.TestModel]
at play.mvc.ActionInvoker.invoke(ActionInvoker.java:231)
at Invocation.HTTP Request(Play!)
aused by: javax.persistence.PersistenceException: org.hibernate.exception.SQLGr
mmarException: could not insert: [models.TestModel]
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityMan
gerImpl.java:1214)
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityMan
gerImpl.java:1147)
at org.hibernate.ejb.AbstractEntityManagerImpl.convert(AbstractEntityMan