Javascript 如何通过提供类名字符串文字来获取类？

如何替换XXX，以便为导入的对应类评估替换？例如，如果名称为“FooService”，则导入的FooService类将在XXX中评估和使用

食品类服务：

export class FooService {}

export class BarService {}

import {FooService} from './FooService'
import {BarService} from './BarService'

export class Container {
    constructor() {
        this._services = new Map();
    }
    getServiceByName(name) {
        if (!this._services.has(name)) {
         // How to replace XXX so that the replacement will 
         // be evaluated for the correspondent class that get imported?
         // e.g. if name is "FooService", then the imported FooService 
         // class will be evaluated and used in XXX.
           this._services.set(name, XXX); 
        }

        return this._service.get(name);
    }
}

分类酒吧服务：

export class FooService {}

export class BarService {}

import {FooService} from './FooService'
import {BarService} from './BarService'

export class Container {
    constructor() {
        this._services = new Map();
    }
    getServiceByName(name) {
        if (!this._services.has(name)) {
         // How to replace XXX so that the replacement will 
         // be evaluated for the correspondent class that get imported?
         // e.g. if name is "FooService", then the imported FooService 
         // class will be evaluated and used in XXX.
           this._services.set(name, XXX); 
        }

        return this._service.get(name);
    }
}

类容器：

export class FooService {}

export class BarService {}

import {FooService} from './FooService'
import {BarService} from './BarService'

export class Container {
    constructor() {
        this._services = new Map();
    }
    getServiceByName(name) {
        if (!this._services.has(name)) {
         // How to replace XXX so that the replacement will 
         // be evaluated for the correspondent class that get imported?
         // e.g. if name is "FooService", then the imported FooService 
         // class will be evaluated and used in XXX.
           this._services.set(name, XXX); 
        }

        return this._service.get(name);
    }
}

---

为什么需要动态地执行此操作？只要用所有的名称和类初始化地图。@Ryan，说得好！IMHO的可能重复除了dirty

eval（name）