Angular2:如何转换我的树模型以适应不同的界面？

Question

假设我有一个看起来像这样的员工类:

export class Employee {

    jobTitle: string;
    firstName: string;
    lastName: string;

    boss: Employee;
    workers: Employee[];

    constructor(jobTitle: string, firstName: string, lastName: string){
        this.jobTitle = jobTitle;
        this.firstName = firstName;
        this.lastName = lastName;
    }

    public addWorker(worker: Employee){
        this.wokers.push(worker);
        worker.boss = this;
    }

}

假设我们的初始Employees看起来像这样:

function getInitialStaff(): Employee {

    let headWaiter: Employee = new Employee( "Head Waiter", "Sarah", "Hilton" );
    headWaiter.addWorker(new Employee("Bus Boy", "Joe", "BeefEater"));
    headWaiter.addWorker(new Employee("Waiter", "Randy", "Rainman"));

    let headCheif: Employee = new Employee( "Head Cheif", "Phil", "Yorstomux" );
    headWaiter.addWorker(new Employee("Dishwasher", "Andy", "Shark"));

    let owner: Employee = new Employee( "Owner", "Boris", "Laughington" );
    owner.addWorker(headWaiter);
    owner.addWorker(headCheif);
    return owner;
}

我们还可以假设有时员工会被雇用或解雇.有点像这样:

window.setTimeout(()=>{
    let someEmployee: Employee = /* Some existing employee */
    someEmployee.addWorker("someTitle", "someName", "someLastName");
}, someAmountOfTime);

---

我想在PrimeNg树中显示员工.

树窗口小部件要求要显示的数据符合treeNode接口,如下所示:

export interface TreeNode {
    label?: string;
    data?: any;
    icon?: any;
    expandedIcon?: any;
    collapsedIcon?: any;
    children?: TreeNode[];
    leaf?: boolean;
}

主要问题:在不更改我的Employee类的情况下,将数据转换为TreeNodes的最佳方法是什么？

---

我正在考虑的事情:

• 当我添加一个新的Employee时,我想保持我的树状态(什么是开放的,什么是选中的,等等)
• 我需要能够轻松检索当前选中的员工.
• 我试图避免维护两个重复的模型.
• 我想以合理但可扩展的方式利用Angular2的变化检测

---

这就是我的EmployeeTree组件目前的样子:

@Component({
    selector: 'employee-tree',
    template: `
        <div>
            <p-tree>
                /* Open Question, I need to transform owner and then tie the resulting TreeNode to the tree */
            </p-tree>
        </div>
    `
    directives: [ Tree ]
})
export class EmployeeTree {
    @Input()
    private owner: Employee;
}

---

选项1 - 创建一个将Employee转换为TreeNode的管道

我想这会让我的模板看起来像这样:

<p-tree [value]="owner | convertToTreeNode"></p-tree>

我会创建一个像这样的管道:

@Pipe({name: 'convertToTreeNode'})
export class ToTreeNode implements PipeTransform {
    transform(employee: Employee): TreeNode {

        let treeNode: TreeNode = {
            label: employee.firstName + " " + employee.lastName,
            data: employee,
            children: []
        };

        employee.workers.forEach(worker => {
            treeNode.children.push(this.transform(worker));
        })

        return treeNode;

    }
}

优点缺点:

• 这利用了角度的变化检测
• 如果这是纯管道,它可能会错过更改
• 如果这是一个不纯的管道,它可能太重了
• 使组件非常干净,并将责任委托给管道
• 添加新项目时,不确定这是否会保留我的gui状态(选择的内容以及打开/关闭的内容)

---

选项2 - 组件内部为树提供单独的结构

@Component({
    selector: 'employee-tree',
    template: `
        <div>
            <p-tree [data]="employeeTree">
            </p-tree>
        </div>
    `
    directives: [ Tree ]
})
export class EmployeeTree {
    @Input()
    private owner: Employee;

    private employeeTree: TreeNode;

    // TODO - Somehow watch the owner for changes (how?) and then repopulate the employeeTree
}

我认为我可以让Employee发出一个在它发生变化时涓涓细流的事件.这可以给我一些东西.

---

一般来说,使用Angular2做这样的事情的最佳方法是什么.

总而言之,我想使用一个小部件,它希望数据的格式与我存储的格式略有不同.将数据按入窗口小部件的标准angular2方法是什么？是否有特殊的角度酱可以帮助我？我应该保留两个单独的型号吗？我的树可能有100个或者可能有1000个物品,所以我有点担心性能.

Answer 1

我会使用 Lodash地图。这是他们的罐装示例：

import * as _ from 'lodash';
var users = [
  { 'user': 'barney' },
  { 'user': 'fred' }
];

// The `_.property` iteratee shorthand.
_.map(users, 'user');
// => ['barney', 'fred']

这是我对此的看法：

import * as _ from 'lodash';
var users = [
  { 'user': 'barney' },
  { 'user': 'fred' }
];

// The `_.property` iteratee shorthand.
_.map(users, 'user');
// => ['barney', 'fred']

let emplData = [
         {  "jobTitle": "Paperboy",
            "firstName": "Fred",
            "lastName":  "Bloggs"
         },
         {  "jobTitle": "Tea boy",
            "firstName": "Charlie",
            "lastName":  "Creamslice"
         }         
    ]
   let resEl = document.getElementById("result");
   resEl.innerText =JSON.stringify(emplData.map(buildEmployeeNamesOnlyAndAddSeqNo));

    function buildEmployeeNamesOnlyAndAddSeqNo({firstName,lastName},index) {
       return {seqNo:index, firstName, lastName}
    }