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根据使用某个映射的输入对象创建新的对象实例

typescript

0

Kirzilla · 技术社区 · 7 年前

假设我们有命令和命令处理程序。每个命令一个CommandHandler。

解决方案应该在代码缩小后工作,并且应该是对重构友好的。

目前我有这种解决方案,我个人不喜欢。

interface Command {}
class ConcreteCommand implements Command {}
class ConcreteCommand2 implements Command {}

interface CommandHandler {}
class ConcreteCommandHandler implements CommandHandler {}
class ConcreteCommand2Handler implements CommandHandler {}

const map = [ //I don't like this kind of map, any other ways?
    [ConcreteCommand, ConcreteCommandHandler],
    [ConcreteCommand2, ConcreteCommand2Handler]
];

function getCommandHandler(command: Command): CommandHandler {
    for(let pair of map) {
        if (command instanceof pair[0]) {
            return new pair[1];
        }
    }
    throw new Error('Unable to find CommandHandler.');
}

console.log( getCommandHandler(new ConcreteCommand()) ); //ConcreteCommandHandler

This code at TypeScript Playground.

0 回复 | 直到 7 年前

1

0

Robbie Milejczak Diego Felipe 7 年前

JavaScript有一个名为Map的数据类型。从 MDN :

使用一个:

const map = new Map()

map.set(ConcreteCommand, ConcreteCommandHandler)

const handler = new (map.get(ConcreteCommand))()

需要注意的是,您的代码无法处理此问题,因为:

getCommandHandler(new ConcreteCommand())

每个实例都将被视为唯一的密钥。您可以使用函数本身,但是: map.set(ConcreteCommand)

2

0

brunnerh 7 年前

Maps 非常适合:

const map = new Map<Command, CommandHandler>([
    [ConcreteCommand, ConcreteCommandHandler],
    [ConcreteCommand2, ConcreteCommand2Handler]
])

function getCommandHandler(command: Command): CommandHandler
{
    if (map.has(command) == false)
        throw new Error('Unable to find CommandHandler.');

    return map.get(command);
}

3

0

Andrey Golubov 7 年前

interface Cmd<T = string> { _T: T  }
interface CmdHandler<T = string> { _T: T }

declare class CmdOne implements Cmd<'CmdOne'> { _T: 'CmdOne' }
declare class CmdTwo implements Cmd<'CmdTwo'> { _T: 'CmdTwo' }
declare class CmdThree implements Cmd<'CmdThree'> { _T: 'CmdThree' }

declare class CmdHandlerOne implements CmdHandler<'CmdOne'> { _T: 'CmdOne' }
declare class CmdHandlerTwo implements CmdHandler<'CmdTwo'> { _T: 'CmdTwo' }
//declare class CmdHandlerThree implements CmdHandler<'CmdThree'> { _T: 'CmdThree' }

const cmds = [
    [CmdOne, CmdHandlerOne],
    [CmdTwo, CmdHandlerTwo],
    //[CmdThree, CmdHandlerThree],
];

export function CmdFactory<T>(cmd: Cmd<T>): CmdHandler<T> {
    const handle = cmds.find(pair => cmd instanceof pair[0])[1];
    return new handle as any; /// not good!
}

CmdFactory(new CmdOne)
CmdFactory(new CmdThree) /// Need error

playground