Question

我使用打字稿并做出反应。
使用项目循环 ISech 的 sampleQuery、dammyQuery、foodQuery 和 groupQuery。
我在 save[i].query 中收到以下错误。

tsError

Element implicitly has an 'any' type because expression of type 'string' can't be used to index type 'ISearch'.
  No index signature with a parameter of type 'string' was found on type 'ISearch'.

interface ISearch {
  test: string；
  sampleQuery: { id: number; query: string };
  dammyQuery: { id: number; query: string };
  foodQuery: { id: number; query: string };
  groupQuery: { id: number; query: string };
}

const searchText = (save: ISearch) => {
  const items = ['sampleQuery', 'dammyQuery', 'foodQuery', 'groupQuery'];

  let text = '';

  items.map((i) => {
    if (save[i].query != '') {
      text = text.concat(save[i].query + ',');
    }
  });
  return text;
};

Answer 1

只是

const items = ['sampleQuery', 'dammyQuery', 'foodQuery', 'groupQuery'] as const;

Answer 2

我认为有一种更优雅的方法可以做到这一点，但这是我的解决方案：

items.map((i) => {
    let key = i as keyof ISearch
    if (save[key].query != '') {
      text = text.concat(save[key].query + ',');
    }
});

这是一个有效的游乐场示例：https://www.typescriptlang.org/play?#code/JYOwLgpgTgZghgYwgAgJIGUJyggFsgbwFgAoZZAZzgFsAHAGwgEUBXaATwC5DlgATbiBbUARtADcyAI5soXSmCigA5sgC+40uT41q7Vh24FeA5ENETps+RUUr1msshgB7F3wNyjJwcLFRJGUMFJRBVDS1kZSgXFlpPeWN+XwsAq2DbUPDHNVJSBBcQW0o4ADcIbgwsHHwAXkJIqjpGBO9k5AAGABp0r2QAciaGZmt+9S7InWo9Vp52gEYeoL7+qZnR8cjXd1mk0wAmJetufu2PDbUJp2jY+OO50wBmI+D+m7iEsdySb7ySfMKxWAkGoFGQ9QA2oMaMNPj1Vrp9KN4Wc4QN3ncOP0ALqOUiMMDISAAD0J9X6-TxJGBEFBADpqHBaAAKZnAbiZFQASnBAD4Gk4CcgANYQdjg3jIOBg0XsFwwNCYbB4RzkYAK5lUcoQ2XYunLcUAQnJ-R5xCc5BJZKJEFJdIKIAQcDAmrKEB1Yr1BuQAGoBl1Tar1KQ1FyqQ6KC5GHT6C5lMyrWHSEA

Answer 3

试试：

type TItemKey = 'sampleQuery' | 'dammyQuery' | 'foodQuery' | 'groupQuery';
type TSearch = {
  [key in TItemKey]: { id: number; query: string };
} & { test: string }

const searchText = (save: TSearch) => {
  const items: TItemKey[] = ['sampleQuery', 'dammyQuery', 'foodQuery', 'groupQuery'];

  let text = '';

  items.map((i) => {
    if (save[i].query != '') {
      text = text.concat(save[i].query + ',');
    }
  });
  return text;
};

typescript 数组处理中的类型不匹配

3 个答案: