Question

给出一个由对象联合组成的数组，我试图找出一种通过Flow提取其中的两种唯一类型的方法，而不必使它们保持最新状态。

这是我的较大类型，包含两个不同对象的联合数组：

type Something = {
    attachments: [
        {|
            // TWhatever should only match this shape
            __typename: 'whatever',
            uniqueToWhatever: boolean
        |},
        {|
            __typename: 'foo',
            uniqueToFoo: boolean
        |}
    ]
}

提取联合元素类型相对简单：

type TAttachments = $PropertyType<Something, 'attachments'>;
type TAttachment = $ElementType<TAttachments, number>;

我想做的是将每种不同类型的对象提取到其自己的类型中，以便使这两个语句成为真：

// Should fail, this is not of type TWhatever
const failingCase: TWhatever = {
    __typename: 'foo',
    uniqueToFoo: true
}

// Should work, matches TWhatever
const successfulCase: TWhatever = {
    __typename: 'whatever',
    uniqueToWhatever: true
}

为此，我尝试使用$<Call>，从the docs for $<Call>的角度来看，我认为这是可能的，但我还没有弄清楚。

下面是我的尝试，但是流程告诉我TWhatever的类型仍然与此处以前相同。

function whatevers(
  attachments: TAttachments,
  type: string
) {
    return attachments.filter(a => a.__typename === type);
}

type TWhatevers = $Call<typeof whatevers, TAttachments, 'whatever'>
type TWhatever = $ElementType<TWhatevers, number>;

函数whatevers已经在逻辑上仅返回任何类型的元素，但我认为我没有正确地指示流。

类似地，我在没有type的{{1}}参数的情况下尝试了此操作

whatevers()

这也不完全正确。

我是在尝试将Flow和function whatevers( attachments: TAttachments ) { return attachments.filter(a => a.__typename === 'whatever'); } type TWhatevers = $Call<typeof whatevers, TAttachments> type TWhatever = $ElementType<TWhatevers, number>; // Should fail, this is not of type TWhatever const failingCase: TWhatever = { __typename: 'foo', uniqueToFoo: true } // Should work, matches TWhatever const successfulCase: TWhatever = { __typename: 'whatever', uniqueToWhatever: true }推到超出其功能极限吗？还是我刚开始使用？

从唯一可识别对象的数组中提取联合类型

0 个答案: