GRAILS：找到一个自我引用的一个一对多的关系

Question

Grails中，

如何将一个发现在一个一对多的关系，例如，所有的孩子，

class Employee { 
    static hasMany = [ subordinates: Employee ] 
    static belongsTo = [ manager: Employee ] 
} 

所有儿童使用单个管理者，如何获得所有下属的下属（如遍历对象图）？

Answer 1

如果您不想修改域，则递归闭包工作。否则，你可能在这个例子中一个短暂属性添加到Employee域类像allSubordinates：

class Employee { 
    String name 
    static hasMany = [ subordinates: Employee ] 
    static belongsTo = [ manager: Employee ] 
    static transients = ['allSubordinates'] 
    def getAllSubordinates() { 
     return subordinates ? subordinates*.allSubordinates.flatten() + subordinates : [] 
    } 
} 

这是一个集成测试，看看它在行动：

import grails.test.* 

class EmployeeTests extends GrailsUnitTestCase { 
    Employee ceo 
    Employee middleManager1, middleManager2 
    Employee e1, e2, e3, e4, e5, e6 

    protected void setUp() { 
     super.setUp() 
     ceo = new Employee(name:"CEO") 
      middleManager1 = new Employee(name:"Middle Manager 1") 
       e1 = new Employee(name:"e1") 
       e2 = new Employee(name:"e2") 
       e3 = new Employee(name:"e3") 
      middleManager2 = new Employee(name:"Middle Manager 2") 
       e4 = new Employee(name:"e4") 
       e5 = new Employee(name:"e5") 
       e6 = new Employee(name:"e6") 

     ceo.subordinates = [middleManager1, middleManager2] 
     middleManager1.subordinates = [e1,e2,e3] 
     middleManager2.subordinates = [e4,e5,e6] 
     assert ceo.save() 
    } 

    void testAllSubordinates() { 
     def topLevelManager = Employee.get(ceo.id) 
     assertNotNull(topLevelManager); 
     assertEquals(8, topLevelManager.allSubordinates?.size()) 
    } 
} 

Answer 2

//Make a recursive closure 
def printAll 

printAll = { emp -> 
    subordinates.each { 
         println it 
         printAll emp 
         } 
}