如何让Inquirer等待函数执行完毕后再发起新提问

我用以下代码循环调用Inquirer：

function start() {
  let stop = 0; 
  const askQuestion = () => {
    inquirer
      .prompt([
        {
          type: "list",
          name: "action",
          message: "WHAT TO DO?",
          choices: ["books", "other", "exit"],
          loop: true,
        },
      ])
      .then((answers) => {
        if (stop == 0) {
          if (answers.action == "books") {
            fetchBooks();
          } else if (answers.action == "exit") {
            stop = 1;
            return;
          }
          askQuestion();
        }
      });
  };
  askQuestion();
}

start();

我的fetchBooks函数包含API调用（books数组来自独立JSON文件）：

function fetchBooks() {
  var len = 0;
  var booksAll = [];
  books.forEach((book) => {
    var bookname = APICALL(book.id).then((bxk) => {
      var b = { name: book.name, short: bxk.text };
      booksAll.push(b);
      len++;
      if (len === books.length) {
        console.log(booksAll);
      }
    });
  });
}

（无需关注fetchBooks的正确性，它本身运行正常）

当前问题：选择“books”时，fetchBooks的执行与Inquirer的新提问同时进行，输出重叠。需要让Inquirer等待fetchBooks执行完毕并输出结果后，再发起新的提问。

核心是让fetchBooks返回Promise，通过异步链式调用或async/await，确保它执行完成后再触发下一次提问。

1. 修改fetchBooks，让它返回Promise #

方式一：用Promise.all简化多API调用逻辑（推荐） #

function fetchBooks() {
  // 用Promise.all等待所有API调用并行完成
  return Promise.all(books.map(book => {
    return APICALL(book.id).then(bxk => {
      return { name: book.name, short: bxk.text };
    });
  })).then(booksAll => {
    console.log(booksAll);
    return booksAll; // 可选：返回结果供后续逻辑使用
  });
}

方式二：给原有逻辑套Promise外壳 #

如果不想改动原有的计数逻辑，可以直接用Promise包裹：

function fetchBooks() {
  return new Promise((resolve) => {
    var len = 0;
    var booksAll = [];
    books.forEach((book) => {
      APICALL(book.id).then((bxk) => {
        var b = { name: book.name, short: bxk.text };
        booksAll.push(b);
        len++;
        if (len === books.length) {
          console.log(booksAll);
          resolve(booksAll); // 执行完成后触发resolve
        }
      });
    });
  });
}

2. 修改Inquirer逻辑，等待fetchBooks完成后再提问 #

方式一：链式调用then #

function start() {
  let stop = 0; 
  const askQuestion = () => {
    inquirer
      .prompt([
        {
          type: "list",
          name: "action",
          message: "WHAT TO DO?",
          choices: ["books", "other", "exit"],
          loop: true,
        },
      ])
      .then((answers) => {
        if (stop == 0) {
          if (answers.action == "books") {
            // 等待fetchBooks执行完成后，再发起下一次提问
            fetchBooks().then(() => {
              askQuestion();
            });
          } else if (answers.action == "exit") {
            stop = 1;
            return;
          } else {
            // 其他操作直接继续提问
            askQuestion();
          }
        }
      });
  };
  askQuestion();
}

start();

方式二：用async/await简化代码 #

function start() {
  let stop = 0; 
  // 把askQuestion改成async函数
  const askQuestion = async () => {
    const answers = await inquirer.prompt([
      {
        type: "list",
        name: "action",
        message: "WHAT TO DO?",
        choices: ["books", "other", "exit"],
        loop: true,
      },
    ]);
    if (stop == 0) {
      if (answers.action == "books") {
        await fetchBooks(); // 等待fetchBooks完成
        askQuestion();
      } else if (answers.action == "exit") {
        stop = 1;
        return;
      } else {
        askQuestion();
      }
    }
  };
  askQuestion();
}

start();

内容的提问来源于stack exchange，提问作者M. Uslu