为 DRY 代码重构 Node.JS 回调答案

【问题标题】：Refactor Node.JS callbacks for DRY code为 DRY 代码重构 Node.JS 回调
【发布时间】：2016-05-15 11:49:27
【问题描述】：

ifs 中的两个 find 调用都有以函数开头的回调（例如，文档）。将它重构为 DRYer 的干净方法是什么？谢谢。

    if (connection_id == null) {
        id_connectionsCollection.find({}, {}, function (e, docs) {
            if (e) {
                return callback(e);
            }
            var connectionDetails = null;
            if (docs == null || docs.length == 0) {//if no connections found, use default from config
                connectionDetails = defaultConnectionDetails
            }
            else {

                connectionDetails = docs[0];//just get the first one
            }

            return callback(null, connectionDetails);
        });

    }
    else {
        id_connectionsCollection.find({name: connection_id}, {sort: {updated_at: -1}}, function (e, docs) {
            if (e) {
                return callback(e);
            }
            var connectionDetails = null;
            if (docs == null || docs.length == 0) {
                connectionDetails = defaultConnectionDetails;
            }
            else {

                connectionDetails = docs[0];//just get the first one
            }
            return callback(null, connectionDetails);
        });
    }

【问题讨论】：

标签： javascript node.js refactoring dry

【解决方案1】：

DRY 代码最明显的方法是将回调提取到命名函数，该函数可以作为回调传递给 find 方法的最后一个参数：

 // Can probably think of a better name here...
 doCallback = function(e, docs) {
      if (e)
        return callback(e);

      var connectionDetails = null;

      if (docs == null || docs.length == 0)
        connectionDetails = defaultConnectionDetails;
      else 
        connectionDetails = docs[0];//just get the first one

      return callback(null, connectionDetails);
}

if (connection_id == null)
  id_connectionsCollection.find({}, {}, doCallback);
else 
  id_connectionsCollection.find({name: connection_id}, {sort: {updated_at: -1}}, doCallback);

【讨论】：