Playframework 如何使用hibernate,jpa从数据库中检索数据
我正在使用play-2.1.3框架。我想从数据库中检索数据。我正在使用eclipse、hibernate和postgreSQL 当我想用entity mgr从数据库中检索数据时,它给了我一个错误,我找不到答案。。。请帮帮我 ----------------------日志文件中出现错误--------------------------Playframework 如何使用hibernate,jpa从数据库中检索数据,playframework,Playframework,我正在使用play-2.1.3框架。我想从数据库中检索数据。我正在使用eclipse、hibernate和postgreSQL 当我想用entity mgr从数据库中检索数据时,它给了我一个错误,我找不到答案。。。请帮帮我 ----------------------日志文件中出现错误-------------------------- 2013-10-06 19:08:53,729 - [ERROR] - from org.hibernate.engine.jdbc.spi.SqlExcept
2013-10-06 19:08:53,729 - [ERROR] - from org.hibernate.engine.jdbc.spi.SqlExceptionHelper in play-akka.actor.default-dispatcher-3
ERROR: column client0_._ebean_intercept does not exists
Position: 31
我的实体类:
package models;
import play.db.ebean.*;
import play.data.validation.*;
import javax.persistence.*;
@Entity
@Table(name="clients")
public class Client extends Model {
/**
*
*/
private static final long serialVersionUID = 1L;
@Id
@Constraints.Min(10)
public Long id;
@Constraints.Required
public String username;
@Constraints.Required
public String email;
@Constraints.Required
public String password;
@Constraints.Required
public String passwordsignup_confirm;
public boolean online;
public boolean getOnline() {
return online;
}
public void setOnline(boolean online) {
this.online = online;
}
public String getUsername() {
return username;
}
public void setUsername(String username) {
this.username = username;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
public String getConfirm() {
return passwordsignup_confirm;
}
public void setConfirm(String confirm) {
this.passwordsignup_confirm= confirm;
}
}
------------------------------在我的控制器类中---------
------------------------------------------persistence.xml
<?xml version="1.0" encoding="UTF-8"?>
<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="manager" 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.H2Dialect"/>
</properties>
</persistence-unit>
</persistence>
您必须使用配置文件(application.conf)来创建数据库
enter code here
# Database configuration
# ~~~~~
db.default.driver=org.postgresql.Driver
db.default.url="jdbc:postgresql://localhost:5432/nameof_you_DB"
db.default.user=your_user_name
db.default.password=you_password
您必须使用配置文件(application.conf)来创建数据库
enter code here
# Database configuration
# ~~~~~
db.default.driver=org.postgresql.Driver
db.default.url="jdbc:postgresql://localhost:5432/nameof_you_DB"
db.default.user=your_user_name
db.default.password=you_password
我们面临着同样的问题..任何线索/运气..我们面临着同样的问题..任何线索/运气..你必须使用u play项目的配置文件(application.conf)连接u DB。你必须使用u play项目的配置文件(application.conf)连接u DB。
db.default.driver=org.postgresql.Driver
db.default.url="jdbc:postgresql:postgres"
db.default.user=postgres
db.default.password="root"
db.default.jndiName=DefaultDS
enter code here
# Database configuration
# ~~~~~
db.default.driver=org.postgresql.Driver
db.default.url="jdbc:postgresql://localhost:5432/nameof_you_DB"
db.default.user=your_user_name
db.default.password=you_password