异步/等待,"未解析的变量或类型等待"

Question

    let usersInAuth0 = [];    
    for (let i = 0; i < 5; i++) {
      const usersInAuth0InPage = await getUsersFromAuth0(i); // error in WebStorm
      if (_.isEmpty(usersInAuth0InPage)){
        return res.status(200).send(usersInAuth0);
      } else {
        usersInAuth0.push(usersInAuth0InPage);
      }
    }

    const getUsersFromAuth0 = async (page) => {
      return await management.getUsers({per_page:100, page: page})
    }

试图使用async/await.从auth0管理API,你每页只能获得100个用户,所以我要多次请求,直到我拥有所有用户.

WebStorm发出警告(第3行):"未解析的变量或类型等待".

在这种情况下,我怎么想使用async/await？

Answer 1

为了使用await,它内部的功能需要async.例如:

(async () => {
  let usersInAuth0 = [];    
  for (let i = 0; i < 5; i++) {
    const usersInAuth0InPage = await getUsersFromAuth0(i); // error in webstorm
    if (_.isEmpty(usersInAuth0InPage)){
      return res.status(200).send(usersInAuth0);
    } else {
      usersInAuth0.push(usersInAuth0InPage);
    }
  }

  const getUsersFromAuth0 = page => management.getUsers({per_page:100, page: page});
})();

https://codereviewvideos.com/blog/unresolved-variable-type-await/

请注意,由于management.getUsers返回Promise,您不需要进行getUsersFromAuth0异步 - 只需返回promise await及其分辨率更高.

因为async函数返回promises,所以你需要确保消耗整个代码块的任何东西能够处理它的返回值是一个promise.(您return res.status(200).send(usersInAuth0);实际上不会返回该值 - 如果条件内部有条件,函数将返回一个解析为该值的promise )