小编Jug*_*gal的帖子

我创建了两个具有多对多关系的bean用户和VirtualDomain

@Entity
@Table(name = "tblUser")
public class User implements Serializable {
    private Long id;
    private String username;
    private Set<VirtualDomain> virtualdomainset;
@Id
@Column(name = "id")
@GeneratedValue(strategy = GenerationType.AUTO)
public Long getId() {
    return id;
}

public void setId(Long id) {
    this.id = id;
}

@Column(name = "username", length = 50, nullable = false)
public String getUsername() {
    return username;
}

public void setUsername(String username) {
    this.username = username;
}
   @ManyToMany(targetEntity = VirtualDomain.class, cascade = {CascadeType.PERSIST},fetch=FetchType.EAGER)
    @JoinTable(name = "tblUserDomainRel", joinColumns = @JoinColumn(name …

我想为项目创建一个war文件.该代码适用于创建jar文件.当我运行以下蚂蚁时,build.xml它仍然给出消息jar文件BUILD SUCCESSFULL.

<project name="struts-spring" basedir="../" default="all">
    <!-- Project settings -->
    <property name="project.title" value="Sufalam Struts 2 Tutorials"/>
    <property name="project.jar.file" value="struts-spring.jar"/>

      <path id="class.path">

        <fileset dir="lib">

          <include name="**/*.jar"/>

        </fileset>

        <fileset dir="libext">

          <include name="**/*.jar"/>

        </fileset>

      </path>

        <!-- Classpath for Project -->

        <path id="compile.classpath">

            <pathelement path ="lib/commons-beanutils.jar"/>

            <pathelement path ="lib/commons-digester.jar"/>

            <pathelement path ="lib/struts.jar"/>

            <pathelement path ="libext/servlet-api.jar"/>

            <pathelement path ="libext/catalina-ant.jar"/>

            <pathelement path ="classes"/>

            <pathelement path ="${classpath}"/>

        </path>

        <!-- Check timestamp on files -->

        <target name="prepare">

            <tstamp/>
            <copy
                file="src/struts.xml"
                todir="src/classes"/>


        </target>
        <!-- Copy any resource …