[英]Recursively requiring modules in Node
So, let's say I have two classes, each of which requires the other: 因此,假设我有两个类,每个类都需要另一个:
Department.js: Department.js:
const Person = require("./Person");
class Department{
constructor(id){
this.personel = getPersonel(id).map(person => new Person(person));
}
}
Person.js Person.js
const Department = require("./Department");
class Person {
constructor(id){
this.department = new Department(getDeptOfPerson(id));
}
}
const person = new Person(1);
const coworkers = person.department.personel;
Now, this doesn't work, and I can kinda understand why. 现在,这不起作用,我可以理解为什么。 It says "Person" is not a constructor at Department.js.
它说“ Person”不是Department.js的构造函数。 However, if I put both classes in the same file, it works just fine.
但是,如果我将两个类都放在同一个文件中,则效果很好。
So; 所以; my question is, how do I work around this?
我的问题是,我该如何解决? I'd really rather not keep both of these classes in the same file -- is there a better way?
我真的不想将这两个类都放在同一个文件中-有更好的方法吗?
I'm running the latest version of Node. 我正在运行最新版本的Node。
You can separate initializing objects from fetching data: 您可以将初始化对象与获取数据分开:
// types/Department.js
class Department {
constructor(id, personnel) {
this.id = id;
this.personnel = personnel;
}
}
// types/Person.js
class Person {
constructor(id, department) {
this.id = id;
this.department = department;
}
}
// findDepartment.js
const Department = require('./types/Department');
const Person = require('./types/Person');
function findDepartment(id) {
const personnel = getPersonnel(id).map(person => new Person(person));
return new Department(id, personnel);
}
// findPerson.js
const Department = require('./types/Department');
const Person = require('./types/Person');
function findPerson(id) {
const department = getDeptOfPerson(id);
return new Person(id, department);
}
These can even go back on the types ( findDepartment.js
→ Department.js
, Department.findDepartment = function (id) { …
) if you really want. 如果您确实需要,它们甚至可以返回类型(
findDepartment.js
→ Department.js
, Department.findDepartment = function (id) { …
))。
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