在Grails应用程序中扩展Spring Security UI插件

Question

我一直在努力寻找一种正确扩展spring security ui插件的方法。 我想做的事情显然很简单：我有3个由脚本创建的域类：

• 用户
• 作用
• 的UserRole

我想扩展用户域类，以便添加名字，姓氏，国家，dateofBirth，性别和可见性功能，我想出了这个想法：

我创建了其他2个域类，RegisterUser和UserInfo。

代码如下：

class User {

   transient springSecurityService

   String username
   String password
   long phone
   String email
   boolean enabled = true
   boolean accountExpired
   boolean accountLocked
   boolean passwordExpired

   static hasMany = [openIds: OpenID]

   static transients = ['springSecurityService']

   static constraints = {
      username blank: false, unique: true
      password blank: false
      phone unique:true , nullable:false, blank: false, length:10, range: 3200000000..3940000000
      email email:true, nullable:false, blank:false
   }

   static mapping = {
      phone column :'idUser'
      email column : 'email'
      username column : 'username'
      password column: '`password`'
   }

   Set<Role> getAuthorities() {
      UserRole.findAllByUser(this).collect { it.role }
   }

   def beforeInsert() {
      encodePassword()
   }

   def beforeUpdate() {
      if (isDirty('password')) {
         encodePassword()
      }
   }

   protected void encodePassword() {
      password = springSecurityService?.passwordEncoder ? springSecurityService.encodePassword(password) : password
   }
}

class UserInfo implements Comparable {

   String firstname
   String lastname
   String country
   Date date_of_birth
   String gender

   static belongsTo = RegisteredUser

   static mapping = {
      firstname column : 'firstname'
      lastname column : 'surname'
      date_of_birth column : 'date_of_birth'
      country column : 'country'
      gender column : 'gender'
   }

   static constraints = {
      firstname blank:false, matches:"[a-zA-Z]+"
      lastname blank:false, matches:"[a-zA-Z]+"
      country blank:true, nullable:true
      date_of_birth nullable:true, unique:false
      gender blank:true, nullable:true, inList:["Male", "Female"]
   }

   @Override
   public int compareTo(Object o) {
      // Compare two user basing on their phone numbers and username
      RegisteredUser u = (RegisteredUser) o
      if(u.info.phone == this.phone){
         if(u.info.email == this.email){
            return 0;
         }
         return -1;
      }
      return 1;
   }
}

class RegisteredUser {

   User user
   UserInfo info
   boolean visible

   //Not related to spring security
   static hasOne = UserInfo
   static hasMany = [friendships : Friendship, blocked: Block, trips : Trip]

   static constraints = {

   }

   /**
    * Programmatic exercise...? Do it in a more elegant (groovy) and less java way
    * @return
    */
   static List<RegisteredUser> getContactList_Java(){
      def friendshipList = 
         Friendship.getFriendships(this)
      def contactList = new ArrayList<RegisteredUser>()
      for(Friendship f : friendshipList){
         if(f.user1.equals(this)){
            friendshipList.add(f.user2)
         }else{
            friendshipList.add(f.user1)
         }
      }
   }
   /**
    * Programmatic exercise?... Done! ;D
    * @return
    */
   static List<RegisteredUser> getContactList(){
      Friendship.getFriendships(){((Friendship)it).otherUser(this)}.collect()
   }

   /**
    * 
    * method user_exists
    * Parameters:
    * @param user
    * @return
    * method type:
    * Dynamically chosen
    */
   static user_exists(User u){
      if(User.where{
         username == u.username}.count() > 0){
         true
      }
      false
   }

   /**
    * 
    * method findByPhone
    * Parameters:
    * @param number
    * @return
    * method type:
    * Dynamically chosen
    */
   def findByPhone(int number){
      User.findByPhone(number)
   }

   /**
    * 
    * method findBySomeInfo
    * Parameters:
    * @param infos
    * @return
    * method type:
    * Dynamically chosen
    */
   def findByMail(String mail){
      User.findByEmail(mail)
   }

   /**
    * 
    * method findByName
    * Parameters:
    * @param firstname
    * @return
    * method type:
    * Dynamically chosen
    */
   def findByName(String firstname){
      UserInfo.findByFirstname(firstname)
   }

   /**
    * 
    * method findSentRequests
    * Parameters:
    * @return
    * method type:
    * Dynamically chosen
    */
   def findSentRequests(){
      Request.findBySender(this)
   }

   /**
    * 
    * method findReceivedRequest
    * Parameters:
    * @return
    * method type:
    * Dynamically chosen
    */
   def findReceivedRequest(){
      Request.findByReceiver(this)
   }

   /**
    * 
    * Method findSentActiveRequest
    * Parameters:
    * @return 
    * Method type:
    * Dynamically defined (Default)
    */
   def findSentActiveRequest(){
      Request.findBySenderAndStatus(this, util.Status.ACTIVE)
   }
}

如果我通过bootstrap实例化注册用户，则应用程序根本看不到用户。

我看不出我明显犯的错误。

Answer 1

我建议你删除RegisteredUser类。我认为一个好的解决方案可能是一对一的关系中有一个User Class和Profile Class：

class User{

transient springSecurityService

   String username
   String password
   long phone
   String email
   boolean enabled = true
   boolean accountExpired
   boolean accountLocked
   boolean passwordExpired

   ...

   static hasOne = [profile: UserProfile]

   ...
}


class UserProfile {

   User user
   String firstname
   String lastname
   String country
   Date date_of_birth
   String gender

   ...

}