javascript - 无法通过递归调用 node.js 中的函数从 promisese 获得响应

Question

使用 node.js Q promises 库 我怀疑递归调用函数是否有用。我尝试了两种方法来解决我的问题。在第一种方法中，我尝试直接从递归函数返回，而在第二种方法中，我尝试使用 Q promise 库。

这是我调用该递归函数并期待此处响应的代码。

function getSearchedUserInTree(userId,body){
    var deferred = Q.defer();
    var loggedInUserId = userId;    
    var searchUserIdSql =  "SELECT rc.user_id,CAST(b.node_id AS CHAR) AS node_id FROM refer_codes AS rc LEFT JOIN btrees AS b ON b.user_id=rc.user_id WHERE rc.refer_code='"+body.refer_code+"'";
    connection.query(searchUserIdSql, function (err, result,field) {        

        var getLoggedInNode = "SELECT CAST(b.node_id AS CHAR) AS node_id FROM btrees as b WHERE b.user_id='"+loggedInUserId+"'";
        connection.query(getLoggedInNode, function (err, loggedResult,field) {
            helper.trackRelation(loggedResult[0].node_id,result[0].node_id,function(err,relationResult){
                if(err) throw err;
                deferred.resolve(relationResult);
            });
        });        
    });

    return deferred.promise;
}

从“trackRelation()”函数获得返回的第一种方法。

function trackRelation(parentNode,childNode){
  var deferred = Q.defer(); 
  var childNodeId = bigInt(""+childNode+"");
  var parentNodeId  = bigInt(""+parentNode+"");
  if(bigInt(childNodeId).compare(parentNodeId) == 0){        
    var childSql = "SELECT * FROM users WHERE id='"+childNodeId.toString()+"'";
    connection.query(childSql,function(err,packResult,field){
      console.log(packResult);
      deferred.resolve({status:200,message:'success',packResult});
    });
  }else{
    /* Detact position */
    if(bigInt(childNodeId).mod(2) == 0){
      var tempParentNodeId = bigInt(childNodeId).divide(2);
    }else{
      var subResult = bigInt(childNodeId).subtract(1);
      var tempParentNodeId = subResult.divide(2);
    } 

    /** If reached at top but parent not found **/
    if(bigInt(tempParentNodeId).compare(1) == 0){
      var childSql = "SELECT * FROM users WHERE id='"+tempParentNodeId.toString()+"'";
      connection.query(childSql,function(err,packResult,field){
        console.log(packResult);
        deferred.resolve({status:200,message:'success',packResult});        
      });
    }else{
      return trackRelation(parentNodeId,tempParentNodeId.toString());
    }

  }
  return deferred.promise;
}

从递归函数中获取返回值的第二种方法，无需 promise 直接返回值。

function trackRelation(parentNode,childNode){      
      var deferred = Q.defer();
      var childNodeId = bigInt(""+childNode+"");
      var parentNodeId  = bigInt(""+parentNode+"");      
      if(bigInt(childNodeId).compare(parentNodeId) == 0)
      {               
        deferred.resolve({status:200,message:'success',topUserId:childNodeId.toString()});
        return childNodeId.toString();
      }else
      {
        /* Detact position */
        if(bigInt(childNodeId).mod(2) == 0){
          var tempParentNodeId = bigInt(childNodeId).divide(2);
        }else
        {
          var subResult = bigInt(childNodeId).subtract(1);
          var tempParentNodeId = subResult.divide(2);
        }

        /** If reached at top but parent not found **/
        if(bigInt(tempParentNodeId).compare(1) == 0){          
          return tempParentNodeId.toString();
        }else
        {
          return trackRelation(parentNodeId,tempParentNodeId.toString());
        }
      }
      return deferred.promise;
    }

我没有从这两种方法中得到任何回应。我试图在控制台中打印我需要的值作为响应，我得到的值与期望值相同，但无法获得该值作为响应。

Answer 1

您无法像从回调中取回那样从 promise 中取回值。

您的方法 trackRelation 返回一个 promise ，其值应该使用 .then() 方法访问。

例子:

trackRelation(args).then(function(res){
     //when the promise resolves, this method is called
}, function(err){
     //when the promise rejects, this method is called
});

查看 API 引用页 here .