GRAILS:以自我引用的一对多关系查找所有孩子

Question

在grails中,

如何在一对多关系中找到所有孩子,例如,

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

使用单个管理器,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,则递归闭包有效.否则,您可以向Employee域类添加一个瞬态属性allSubordinates,如下例所示:

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,可以看到它的实际效果:

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())
    }
}