struts2的休眠映射

Question

我正在尝试使用hibernate和struts2构建登录页面。 我的设计如下。 每个登录用户都有一个角色。 许多用户可以具有相同的角色。
所以我的课是：
User.java

@Entity
@Table(name = "user", catalog = "ciner")
public class User implements java.io.Serializable {

    private Integer userId;
    private Role role;
    private String loginId;
    private String password;
    private String firstName;
    private String lastName;

    public User() {
    }

    public User(Role role, String loginId, String password, String firstName, String lastName) {
        this.role = role;
        this.loginId = loginId;
        this.password = password;
        this.firstName = firstName;
        this.lastName = lastName;
    }

    @Id
    @GeneratedValue(strategy = IDENTITY)
    @Column(name = "user_id")
    public Integer getUserId() {
        return this.userId;
    }

    public void setUserId(Integer userId) {
        this.userId = userId;
    }

    @Column(name = "password", nullable = false)
    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "role_id")
    public Role getRole() {
        return this.role;
    }

    public void setRole(Role role) {
        this.role = role;
    }

    @Column(name = "login_id", nullable = false, length = 7)
    public String getLoginId() {
        return this.loginId;
    }

    public void setLoginId(String loginId) {
        this.loginId = loginId;
    }

    @Column(name = "first_name", nullable = false, length = 50)
    public String getFirstName() {
        return this.firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    @Column(name = "last_name", nullable = false, length = 50)
    public String getLastName() {
        return this.lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    @Override
    public String toString() {
        return "User{" + "userId=" + userId + "role=" + role + "loginId="
                + loginId + "firstName=" + firstName + "lastName=" + lastName
                + '}';
    }

}

角色.java

@Entity
@Table(name = "role", catalog = "ciner")
public class Role implements java.io.Serializable {

    private Integer roleId;
    private String roleDescription;
    private List users;

    public Role() {
    }

    public Role(String roleDescription) {
        this.roleDescription = roleDescription;
    }

    public Role(String roleDescription, List users) {
        this.roleDescription = roleDescription;
        this.users = users;
    }

    @Id
    @GeneratedValue(strategy = IDENTITY)
    @Column(name = "role_id", unique = true, nullable = false)
    public Integer getRoleId() {
        return this.roleId;
    }

    public void setRoleId(Integer roleId) {
        this.roleId = roleId;
    }

    @Column(name = "role_description", nullable = false, length = 100)
    public String getRoleDescription() {
        return this.roleDescription;
    }

    public void setRoleDescription(String roleDescription) {
        this.roleDescription = roleDescription;
    }

    // @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY, mappedBy =
    // "role")
    @OneToMany(fetch = FetchType.LAZY, mappedBy = "role", targetEntity = User.class)
    public List getUsers() {
        return this.users;
    }

    public void setUsers(List users) {
        this.users = users;
    }

    @Override
    public String toString() {
        return "Role{" + "roleId=" + roleId + "roleDescription="
                + roleDescription + "users=" + users + '}';
    }
}

如果有效，它将返回1个用户。 但是由于该方法使该用户返回，因此我使用return users.get(0);

堆

java.lang.StackOverflowError
    java.lang.Integer.toString(Unknown Source)
    java.lang.Integer.toString(Unknown Source)
    java.lang.String.valueOf(Unknown Source)
    java.lang.Integer.toString(Unknown Source)
    java.lang.String.valueOf(Unknown Source)
    java.lang.StringBuilder.append(Unknown Source)
    com.inhis.model.Role.toString(Role.java:70)
    sun.reflect.GeneratedMethodAccessor293.invoke(Unknown Source)
    sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source)
    java.lang.reflect.Method.invoke(Unknown Source)
    org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer.invoke(JavassistLazyInitializer.java:197)
    com.inhis.model.Role_$$_javassist_6.toString(Role_$$_javassist_6.java)
    java.lang.String.valueOf(Unknown Source)
    java.lang.StringBuilder.append(Unknown Source)
    com.inhis.model.User.toString(User.java:97)
    java.lang.String.valueOf(Unknown Source)
    java.lang.StringBuilder.append(Unknown Source)
    java.util.AbstractCollection.toString(Unknown Source)
    org.hibernate.collection.PersistentBag.toString(PersistentBag.java:507)

说，角色是
1，未经授权
2，经理
3，用户
我究竟做错了什么？ 另外，我想为他们分配默认角色1。我不知道这一点。 因此，我现在将其存储为null。 我该如何实现？

Answer 1

您创建了一个无限循环。 如果调用角色的toString方法，它将为每个用户调用toString方法，这将再次调用角色的toString。 一个可能的解决方案是将其从角色类中删除：

"users=" + users +