等待 Promise.all 无效

Question

下面是我的 async 函数，用于从数据库读取值并在控制台上记录一些内容。但由于某种原因，它没有发生。

所以，我创建了一组承诺，然后继续等待所有承诺解决，因为承诺将从数据库中读取。但是 Promise.all 上的 await 并没有暂停代码的执行，也没有等待所有承诺解决，而是将控制权传递回调用函数。 我在这里有什么遗漏吗？

async function readValuesFromDB(codes) {
  console.log('trying to read values from DB');
  let dbPromises = [];
  for(let code in codes) {
    let refId = 'some_random_reference_id_generated_from_codes[i]';
    dbPromises.push(MyDb.getPromise(refId)); // getPromise() returns a promise that reads from DB
  }
  console.log('waiting for all promises to resolve...');

  // waiting for all promises to finish 
  await Promise.all(dbPromises).then((allPromises) => {
    for(let i in allPromises) {
      console.log('OK..read from DB');
    }
  });
  console.log('all promises resolved');

}

console.log('calling readValuesFromDB()');
readValuesFromDB();
console.log('finished readValuesFromDB()');

上述调用的输出是：

trying to read values from DB
waiting for all promises to resolve...
finished readValuesFromDB
......
......
OK..read from DB
OK..read from DB
OK..read from DB
...
...
all promises resolved

理想的输出应该在下面（因为我正在等待使用 await 和 Promise.all 解决所有承诺）

trying to read values from DB
waiting for all promises to resolve...
OK..read from DB
OK..read from DB
OK..read from DB
...
...

因此，在控件传递回调用函数后，promise 似乎正在得到解决，并且似乎 await 在这里没有任何效果。

关于为什么 await 在这种情况下无效的任何帮助？

Answer 1

您在这里缺少等待：

(async function() {
  console.log('calling readValuesFromDB()');
  await readValuesFromDB();
  console.log('finished readValuesFromDB()');
})();

Answer 2

行为正常。

readValuesFromDB 是异步运行的，而不是 console.log('finished readValuesFromDB()');

所以控制台在 readValuesFrom DB 解析之前解析“已完成”。

你可以这样做：

async function readValuesFromDB(codes) {
      console.log('trying to read values from DB');
      let dbPromises = [];
      for(let code in codes) {
        let refId = 'some_random_reference_id_generated_from_codes[i]';
        dbPromises.push(MyDb.getPromise(refId)); // getPromise() returns a promise that reads from DB
      }
      console.log('waiting for all promises to resolve...');
    
      // waiting for all promises to finish 
      let allPromises = await Promise.all(dbPromises);
  for(let i in allPromises) {
    console.log('OK..read from DB');          
  }  
  console.log('all promises resolved');
      console.log('finished readValuesFromDB()');
    }
    
    console.log('calling readValuesFromDB()');
    readValuesFromDB();

或者像这样：

async function readValuesFromDB(codes) {
  console.log('trying to read values from DB');
  let dbPromises = [];
  for(let code in codes) {
    let refId = 'some_random_reference_id_generated_from_codes[i]';
    dbPromises.push(MyDb.getPromise(refId)); // getPromise() returns a promise that reads from DB
  }
  console.log('waiting for all promises to resolve...');
  
  // waiting for all promises to finish 
  let allPromises = await Promise.all(dbPromises);
  for(let i in allPromises) {
    console.log('OK..read from DB');          
  }  
  console.log('all promises resolved');
  allPromises = ['a', 'b', 'c']; // Dumb values for example
  return allPromises;
}

console.log('calling readValuesFromDB()');
readValuesFromDB().then(response => {
   console.log('finished readValuesFromDB()');
   console.log(response)
});

Answer 3

您必须通过await 呼叫readValuesFromDB。

console.log('calling readValuesFromDB()');
await readValuesFromDB();
console.log('finished readValuesFromDB()');