Question

我认为解释我的情况的最好方法是使用代码：

interface IPluginSpec {
  name: string;
  state?: any;
}

interface IPluginOpts<PluginSpec extends IPluginSpec> {
  name: PluginSpec['name'];
  // How to require opts.initialState ONLY when PluginSpec['state'] is defined?
  initialState: PluginSpec['state'];
}

function createPlugin<PluginSpec extends IPluginSpec>(
  opts: IPluginOpts<PluginSpec>,
) {
  console.log('create plugin', opts);
}

interface IPluginOne {
  name: 'pluginOne';
  // Ideally state would be omitted here, but I can also live with having to
  // define "state: undefined" in plugins without state
  // state: undefined;
}

// Error: Property 'initialState' is missing in type...
createPlugin<IPluginOne>({
  name: 'pluginOne',
  // How to make initialState NOT required?
  // initialState: undefined,
  // How to make any non-undefined initialState invalid?
  // initialState: 'anything works here',
});

interface IPluginTwo {
  name: 'pluginTwo';
  state: number;
}

createPlugin<IPluginTwo>({
  name: 'pluginTwo',
  initialState: 0,
});

Answer 1

您可以使用条件类型来执行此操作。有了它，您可以测试该属性是否存在，并且可以拥有或不具有额外的属性：

interface IPluginSpec {
  name: string;
  state?: any;
}

type IPluginOpts<PluginSpec extends IPluginSpec> = PluginSpec extends Record<'state', infer State> ? {
  name: PluginSpec['name'];
  initialState: State;
} : {
  name: PluginSpec['name']
}

function createPlugin<PluginSpec extends IPluginSpec>(
  opts: IPluginOpts<PluginSpec>,
) {
  console.log('create plugin', opts);
}

interface IPluginOne {
  name: 'pluginOne';
}

// Ok
createPlugin<IPluginOne>({
  name: 'pluginOne',
  // nothing to add
});

interface IPluginTwo {
  name: 'pluginTwo';
  state: number;
}

createPlugin<IPluginTwo>({
  name: 'pluginTwo',
  initialState: 0,
});

对于更可组合的方法，您可以使用具有公共部分的交集，并且每个交集中的可选部分都是自己有条件的：

interface IPluginSpec {
    name: string;
    state?: any;
    config?: any;
}

type IPluginOpts<PluginSpec extends IPluginSpec> = {
        name: PluginSpec['name']
    }
    & (PluginSpec extends Record<'state', infer State> ? { initialState: State; } : {})
    & (PluginSpec extends Record<'config', infer Config> ? { initialConfig: Config; } : {})

条件类型对调用者非常有用。问题在于，在实现内部，打字稿无法真正说明条件类型（因为未知T）。

最好的解决方案是保留公共签名（带有条件类型）和简化的实现签名（没有条件类型）。这样，您就可以在不给类型断言的情况下实现函数，同时为调用者提供所需的行为：

function createPlugin<PluginSpec extends IPluginSpec>(opts: IPluginOpts<PluginSpec>)
function createPlugin<PluginSpec extends IPluginSpec>(opts: {
    name: string
    initalState: PluginSpec['state'],
    initialConfig: PluginSpec['config'],
}) {
    if (opts.initalState) {
        opts
    }
}

如何基于TypeScript中的其他类型使对象属性为可选？

1 个答案: