Question

以下抽象的TS场景：

[Example] Attena Illusive - 01 [720p].mkv

在控制台记录SELECT ts_rank_cd(to_tsvector('english', name), query, 1) as rank, * FROM tables, plainto_tsquery('Attena Illusive 1') query WHERE to_tsvector('english', name) @@ query ORDER BY rank desc时引发TypeScript错误，因为它显然不存在。

interface EmotionsOkay {
  emotion: string;
  okay: "yap";
}
interface EmotionsNotOkay {
  emotion: string;
}
type UndetereminedEmotion = EmotionsOkay | EmotionsNotOkay;
const areYouOkay = (test: UndetereminedEmotion) => {
  console.log(test.okay ? "happy :D" : "sad D:");
};

即使传递给该方法的测试的类型为test.okay，也很可能存在。

为什么会这样？

Answer 1

此时，TS不知道它是什么类型，可能是EmotionsOkay，一切都会好起来，或者可能是EmotionsNotOkay，并且属性okay将不存在。这就是为什么您会收到此错误。您需要验证它是哪种类型。

您可以使用in关键字进行验证，

interface EmotionsOkay {
  emotion: string;
  okay: "yap";
}
interface EmotionsNotOkay {
  emotion: string;
}
type UndetereminedEmotion = EmotionsOkay | EmotionsNotOkay;
const areYouOkay = (test: UndetereminedEmotion) => {
  console.log('okay' in test && test.okay ? "happy :D" : "sad D:");
};

如果您首先没有属性okay，就不会幸福。

如果您想知道为什么在这种情况下不能使用instanceof，请查看this question

Here is a playground

Answer 2

您可以使用Typescript type assertion（类似于其他语言的投射）。

interface EmotionsOkay {
  emotion: string;
  okay: "yap";
}
interface EmotionsNotOkay {
  emotion: string;
}
type UndetereminedEmotion = EmotionsOkay | EmotionsNotOkay;
const areYouOkay = (test: UndetereminedEmotion) => {
  console.log((test as EmotionsOkay).okay ? "happy :D" : "sad D:");
  // or
  console.log((<EmotionsOkay>test).okay ? "happy :D" : "sad D:");
};

Typescript：组合类型不存在属性？

2 个答案: