递归调用返回承诺的异步函数

Question

我正在尝试递归调用 AWS 的 SNS listEndpointsByPlatformApplication。这将返回前 100 个端点，然后返回 NextToken 中的令牌（如果还有更多要返回）（详细信息：AWS SNS listEndpointsByPlatformApplication）。

这是我尝试过的：

var getEndpoints = function(platformARN, token) {

  return new models.sequelize.Promise(function(resolve, reject) {
    var params = {
      PlatformApplicationArn: platformARNDev
    };
    if (token != null) {
      params['NextToken'] = token;
    }
    sns.listEndpointsByPlatformApplication(params, function(err, data) {
      if (err) {
        return reject(err);
      }
      else {
        endpoints = endpoints.concat(data.Endpoints); //save to global var
        if ('NextToken' in data) {
          //call recursively
          return getEndpoints(platformARN, data.NextToken);
        }
        else {
          console.log('trying to break out!');
          return resolve(true);          
        }
      }
    });
  });
}

我是这样称呼它的：

getEndpoints(platformARNDev, null)
.then(function(ret) {
  console.log('HERE!');
}, function(err) {
  console.log(err);
});

问题是：第一次调用发生，然后递归调用发生，我收到消息trying to break out!，但HERE! 永远不会被调用。我认为我的承诺的回报方式有问题。

感谢指点。

Answer 1

问题是您尝试解决/拒绝部分完成的查询。这是一个带有虚拟服务的完整工作示例。我将数据抓取封装到它自己的递归函数中，并且仅在我完全获取所有数据或偶然发现错误时才解析/拒绝：

// This is the mock of the service. It yields data and token if
// it has more data to show. Otherwise data and null as a token.
var dummyData = [0, 1, 2, 3, 4];
function dummyAsyncCall(token, callback) {
  token = token || 0;
  setTimeout(function() {
    callback({
        dummyDataPart: dummyData[token],
        token: (typeof (dummyData[token]) == 'undefined') ? null : (token + 1)
    });
  });
}

// Here is how you would recursively call it with promises:

function getAllData() {
  //data accumulator is sitting within the function so it doesn't pollute the global namespace.
  var dataSoFar = [];

  function recursiveCall(token, resolve, reject) {
    dummyAsyncCall(token, function(data) {
      if (data.error) {
        reject(data.error);
      }
      if (!data.token) {
        //You don't need to return the resolve/reject result.
        resolve(dataSoFar);
      } else {
        dataSoFar = dataSoFar.concat(data.dummyDataPart);
        recursiveCall(data.token, resolve, reject);
      }
    });
  }

  return new Promise(function(resolve, reject) {
    // Note me passing resolve and reject into the recursive call.
    // I like it this way but you can just store them within the closure for
    // later use
    recursiveCall(null, resolve, reject);
  });
}

//Here is the call to the recursive service.
getAllData().then(function(data) {
  console.log(data);
});

Fiddle with me

Answer 2

那是因为你不需要返回resolve/reject，只需在递归调用完成时调用resolve/reject。粗略的代码如下所示

var getEndpoints = function(platformARN, token) {

  return new models.sequelize.Promise(function(resolve, reject) {
    var params = {
      PlatformApplicationArn: platformARNDev
    };
    if (token != null) {
      params['NextToken'] = token;
    }
    sns.listEndpointsByPlatformApplication(params, function(err, data) {
      if (err) {
        reject(err);
      }
      else {
        endpoints = endpoints.concat(data.Endpoints); //save to global var
        if ('NextToken' in data) {
          //call recursively
          getEndpoints(platformARN, data.NextToken).then(function () {
            resolve(true);
          }).catch(function (err) {
            reject(err);
          });
        }
        else {
          console.log('trying to break out!');
          resolve(true);          
        }
      }
    });
  });
}

（注意：这只是一个粗略的代码，可能有效，也可能无效，但只是给出一个大致的想法）

我在下面添加了一个代码 sn-p 来支持这个概念，并且效果很好，请查看。

i = 0;

$('#output').empty();

function pro() {
  return new Promise(function(resolve, reject) {
    if (i > 3) {
      resolve();
      return;
    }
    window.setTimeout(function() {
      console.log(i);
      $('#output').append(i).append('<br/>');
      i += 1;
      pro().then(function() {
        resolve()
      }).catch(function() {
        reject()
      });
    }, 2000);
  });
}

pro().then(function () { $('#output').append("now here"); })

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

<div id="output"></div>