저는 playframework 2.3.x와 Java를 사용하고 있습니다. 내 프로젝트를 위해 내가 JPA, 연결 작업 벌금을 (내가 생각하는) MySQL과 연결하여 사용하고자하지만 난이 모델 사용자를 생성 할 때이 오류가 :Playframework 2.3.x JAVA + JPA + MySQL
[error] C:\Users\lpiroche\Documents\www\play_framework\play_sample1\modules\core
\app\models\User.java:3: error: package javax.persistence does not exist
[error] import javax.persistence.Column;
[error] ^
[error] C:\Users\lpiroche\Documents\www\play_framework\play_sample1\modules\core
\app\models\User.java:4: error: package javax.persistence does not exist
[error] import javax.persistence.Entity;
[error] ^
[error] C:\Users\lpiroche\Documents\www\play_framework\play_sample1\modules\core
\app\models\User.java:5: error: package javax.persistence does not exist
[error] import javax.persistence.GeneratedValue;
[error] ^
[error] C:\Users\lpiroche\Documents\www\play_framework\play_sample1\modules\core
\app\models\User.java:6: error: package javax.persistence does not exist
[error] import javax.persistence.GenerationType;
[error] ^
[error] C:\Users\lpiroche\Documents\www\play_framework\play_sample1\modules\core
\app\models\User.java:7: error: package javax.persistence does not exist
[error] import javax.persistence.*;
[error]^
[error] C:\Users\lpiroche\Documents\www\play_framework\play_sample1\modules\core
\app\models\User.java:9: error: cannot find symbol
[error] @Entity
[error]^
[error] symbol: class Entity
[error] C:\Users\lpiroche\Documents\www\play_framework\play_sample1\modules\core
\app\models\User.java:10: error: cannot find symbol
[error] @Table(name="users")
연극 문서는 예 가난을 mysql 데이터베이스에 연결하려면 tuto, post를 읽으십시오. 그러나 해결책을 찾지 못합니다. 나는 persistence.xml 파일에서 뭔가를 잊어 버렸다고 생각하지만, 나는 무엇을 모른다.
build.sbt
libraryDependencies ++= Seq(
javaJdbc,
cache,
javaWs,
javaJpa,
"org.hibernate" % "hibernate-entitymanager" % "3.6.9.Final",
"mysql" % "mysql-connector-java" % "5.1.18"
)
application.conf
db.default.driver=com.mysql.jdbc.Driver
db.default.url="jdbc:mysql://localhost/mydb"
db.default.user=root
db.default.password=""
db.default.jndiName=DefaultDS
jpa.default=defaultPersistenceUnit
예에서
<persistence xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd"
version="2.0">
<persistence-unit name="defaultPersistenceUnit" transaction-type="RESOURCE_LOCAL">
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<non-jta-data-source>DefaultDS</non-jta-data-source>
<properties>
<property name="hibernate.dialect" value="org.hibernate.dialect.MySQLDialect" />
<property name="hibernate.hbm2ddl.auto" value="update"/>
<property name="hibernate.show_sql" value="true"></property>
<property name="hibernate.format_sql" value="true"></property>
</properties>
</persistence-unit>
</persistence>
의 persistence.xml 내가 differentes 구성을 찾을 수 :
<property name="hibernate.hbm2ddl.auto" value="update"/>
<property name="hibernate.show_sql" value="true"></property>
<property name="hibernate.format_sql" value="true"></property>
또는
<property name="hibernate.dialect" value="org.hibernate.dialect.MySQLDialect" />
<property name="hibernate.connection.driver_class" value="com.mysql.jdbc.Driver" />
내가 가장 적절한
을, Hibernate는 것을해야 의존성 등 ... ./activator update'를 실행하고 다시 시도하십시오.같은 결과를 얻으면 build.html의 libraryDependencies에''org.hibernate.javax.persistence "%"hibernate-jpa-2.0-api "%"1.0.1.Final "을 추가하고'를 실행하십시오./activator update' – Salem