Question

在以下代码中，expiringContentsAllBU填充在getOnlyContentWhatsNewReport内的异步/等待函数array.map()中，但是expiringContentsAllBU在{{ 1}}函数。

businessUnits.map()

Answer 1

var getOnlyContentWhatsNewReport = Promise.resolve(123);

const getExpiringContents = async () => {
  let businessUnits = [{ id: 1 }, { id: 2 }, { id: 3 }];  
  const expiringContentsAllBU = await Promise.all(businessUnits.map(async (bu) => {
      return getOnlyContentWhatsNewReport.then(respBu => {
        bu.businessUnit = respBu;
        return bu;
      }).catch((err) => {
        console.log('No expiring contents found for WhatsNewSection');
        return null;
      });
   }));
   console.log('result', expiringContentsAllBU);
}

getExpiringContents();

您必须等到映射完成并且所有回调都完成后。 console.log和后续代码块将在映射完成之前执行，所以

const getExpiringContents = async () => {
  let businessUnits = Object.keys(contentFulSpaces);  
  const expiringContentsAllBU = await Promise.all(businessUnits.map(async (bu) => {
      return getOnlyContentWhatsNewReport(bu, (err, result) => { 
        if(result) {
          let expiringContentsByBU = {};
          expiringContentsByBU['businessUnit'] = bu;
          expiringContentsByBU['expiringContents'] = result;
          return JSON.parse(JSON.stringify(expiringContentsByBU);
        } else {
          console.log('No expiring contents found for WhatsNewSection');
          return null;
        }
      })
   }));
   console.log('result', expiringContentsAllBU);
}

Answer 2

由于map不了解异步功能，因此您需要使用异步功能。一个示例是等效的Bluebird Promise.map：

const getExpiringContents = async () => {
  let businessUnits = Object.keys(contentFulSpaces);  

  // Use Promise.map here to convert each businessUnit entry into a record
  let expiringContentsAllBU = await Promise.map(businessUnits, async (bu) => {
    await getOnlyContentWhatsNewReport(bu, (err, result) => { 
      if (!result) {
        console.log('No expiring contents found for WhatsNewSection');
        return;
      }

      let expiringContentsByBU = {};
      expiringContentsByBU['businessUnit'] = bu;
      expiringContentsByBU['expiringContents'] = result;

      return JSON.parse(JSON.stringify(expiringContentsByBU));
    })
  });

  // As this may contain undefined elements, remove those
  expiringContentsAllBU = expiringContentsAllBU.filter(e => e);

  console.log('result', expiringContentsAllBU);
}

如果您使getOnlyContentWhatsNewReport返回应有的承诺而不是使用回调方法，则可以使此代码更加平坦。 async不会等待回调方法，因此请确保还返回承诺，否则此代码将无法正确等待。

完全重构的版本看起来像这样：

const getExpiringContents = async () => {
  let businessUnits = Object.keys(contentFulSpaces);  

  let expiringContentsAllBU = await Promise.map(businessUnits, async businessUnit => {
    let expiringContents = await getOnlyContentWhatsNewReport(businessUnit);

    if (!expiringContents) {
      console.log('No expiring contents found for WhatsNewSection');
      return;
    }

    // Choosing names that match the output means you can use the destructuring operator
    let expiringContentsByBU = {
      businessUnit,
      expiringContents
    };

    // A more formalized "clone" function could help here.
    return JSON.parse(JSON.stringify(expiringContentsByBU));
  });

  // As this may contain undefined elements, remove those
  expiringContentsAllBU = expiringContentsAllBU.filter(e => e);

  console.log('result', expiringContentsAllBU);
}

Answer 3

您可以将.map()更改为for loop。

或者在您的.map()函数中，返回承诺。然后，您可以致电await Promise.all(promiseArray)

Answer 4

使用带有lodash函数的异步来保持理智-

getExpiringContents = async() => {
  let businessUnits = Object.keys(contentFulSpaces);

  let expiringContentsAllBU = await Promise.map(businessUnits, async businessUnit => {
    let expiringContents = await getOnlyContentWhatsNewReport(businessUnit);

    if (_.isEmpty(expiringContents)) {
      console.log('No expiring contents found for WhatsNewSection');
      return;
    }

    // Choosing names that match the output means you can use the destructuring operator
    let expiringContentsByBU = {
      businessUnit,
      expiringContents
    };

    // A more formalized "clone" function could help here.
    return _.cloneDeep(expiringContentsByBU);
  });

  // As this may contain undefined elements, remove those
  expiringContentsAllBU = _.compact(expiringContentsAllBU);

  console.log('result', expiringContentsAllBU);
}

async / await不会将构建的数组返回给父array.map（）

4 个答案: