EclipseLink/MySQL/Glassfish - 即使连接工作正常,表格也不会创建
我想设置一个基本的JPA应用程序,它在mySQL中创建一个简单的表。EclipseLink/MySQL/Glassfish - 即使连接工作正常,表格也不会创建
我使用glassfish 4.1,我创建了连接池/资源并成功从管理面板中ping数据库。
我创建了一个web应用程序项目,其中包含一个简单的实体和src/META-INF文件夹下的persistence.xml。
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1" xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">
<persistence-unit name="DataPersistencePU" transaction-type="JTA">
<jta-data-source>mysqlresource</jta-data-source>
<exclude-unlisted-classes>false</exclude-unlisted-classes>
<properties>
<property name="javax.persistence.schema-generation.database.action" value="create"/>
</properties>
</persistence-unit>
</persistence>
与实体文件,
@Entity
public class DataProvider implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
@Override
public int hashCode() {
int hash = 0;
hash += (id != null ? id.hashCode() : 0);
return hash;
}
@Override
public boolean equals(Object object) {
// TODO: Warning - this method won't work in the case the id fields are not set
if (!(object instanceof DataProvider)) {
return false;
}
DataProvider other = (DataProvider) object;
if ((this.id == null && other.id != null) || (this.id != null && !this.id.equals(other.id))) {
return false;
}
return true;
}
@Override
public String toString() {
return "Entity.DataProvider[ id=" + id + " ]";
}
}
我也试过在/ WEB-INF移动persistence.xml中,仍无法治愈。
Glassfish日志中没有错误。 我只是想让eclipseLink工作并自动生成表格。
暗提到的,我创建一个EntityManager,开始了交易,并坚持一排表:
@WebServlet(name = "TestUsertServlet", urlPatterns = {"/TestUsertServlet"})
@PersistenceContext(
unitName = "DataPersistencePU")
@TransactionAttribute(REQUIRED)
public class TestUsertServlet extends HttpServlet {
@Resource
private javax.transaction.UserTransaction utx;
@PersistenceContext
EntityManager em;
protected void processRequest(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html;charset=UTF-8");
try (PrintWriter out = response.getWriter()) {
/* TODO output your page here. You may use following sample code. */
out.println("<!DOCTYPE html>");
out.println("<html>");
out.println("<head>");
out.println("<title>Servlet TestUsertServlet</title>");
out.println("</head>");
out.println("<body>");
out.println("<h1>Servlet TestUsertServlet at " + request.getContextPath() + "</h1>");
out.println("</body>");
out.println("</html>");
}
}
@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html;charset=UTF-8");
try {
utx.begin();
TestUser testUser = new TestUser();
em.persist(testUser);
utx.commit();
} catch (Exception ex) {
ex.printStackTrace();
}
}
但想知道为什么需要EntityManager的为了要创建的表。 我认为EM只能使用持久性上下文,实体与数据库相关的生命周期,这意味着对象表示记录到表,而不是表本身,对吗?
EclipseLink懒惰地部署持久性单元以避免在类路径中找到许多可能会被忽略的开销。如果要将其更改为预先部署,请参见https://www.eclipse.org/eclipselink/documentation/2.4/jpa/extensions/p_deploy_on_startup.htm#delayonstartup。 – Chris
好的问这里明显。你是否在应用程序的某个地方创建了“EntityManager”? – Ranjeet