如何删除重复结果并合并在 mySQL 查询中具有相似属性的对象？

Posted 2023-02-24

【中文标题】如何删除重复结果并合并在 mySQL 查询中具有相似属性的对象？

【英文标题】：How can I remove duplicate results and merge objects that have a similar property on a mySQL query?

【发布时间】：2017-08-13 06:44:18

【问题描述】：

我正在使用带有 node.js 的 mysql 库来创建这个查询：

SELECT vols.id_vol, vols.id_place, vols.id_vol_type, vols.id_user_creator, vols.name, vols.desc, vols.date_creation, vols.date_begin, vols.date_end, vols.duration, ' +
            ' vols.active, vols.start_time, vols.end_time, vols.insurance, vols.deleted, users.id_user, users.login, users.verified, users.photo_url, comments.message, comments.id_user, place.id_place, place.name AS placeName, place.lat, place.long' +
            ' FROM vols INNER JOIN users ON vols.id_user_creator = users.id_user INNER JOIN place ON vols.id_place = place.id_place LEFT JOIN comments ON comments.id_vol = vols.id_vol WHERE vols.deleted = 0

返回这个：

  "success": true,
  "vols": [
    
      "vol": 
        "id_vol": 1,
        "id_place": 1,
        "id_vol_type": 2,
        "id_user_creator": 1,
      ,
      "users": 
        "id_user": 1,
        "login": "",
        "verified": 0,
      ,
      "comments": 
        "message": "Muito fixe",
        "id_user": 3
      ,

      "vol": 
        "id_vol": 1,
        "id_place": 1,
        "id_vol_type": 2,
        "id_user_creator": 1,
      ,
      "users": 
        "id_user": 1,
        "login": "",
        "verified": 0,
      ,
      "comments": 
        "message": "ola",
        "id_user": 3
      

当前，它正在为具有该特定 vol ID 的每条评论重复“vol”。是否可以停止显示重复项并将 cmets 合并到单个数组上？

【参考方案1】：

这是不可能的，我的意思是你可以用 group_concat 函数做一些事情，但是会很麻烦。

在类似情况下我通常会做的是：

查询出所有相关卷 提取您要查看的所有用户ID、地点ID、评论ID。（在运行所有结果以提取此数据时，最好为每个结果创建一个对象，并将它们放入索引数组中（对象）对于每个可索引参数“用户、地点、评论”） 进行 3 个单独的查询，每个查询一个（用户、地点、cmets），您只选择相关元素（使用 WHERE id IN (1,5,7,8,10) 过滤器） 遍历每个查询的结果，并使用索引数组（对象）将它们插入到 vol 对象中

在 javascript 中类似于（抱歉不知道 node.js，所以我没有一个精确的解决方案

var results = db.queryResults('SELECT vols.id_vol, vols.id_place, vols.id_vol_type, vols.id_user_creator, vols.name, vols.desc, vols.date_creation, vols.date_begin, vols.date_end, vols.duration, vols.active, vols.start_time, vols.end_time, vols.insurance, vols.deleted FROM vols WHERE vols.deleted = 0');
var lookupByUserID = ;
var lookupByPlaceID = ;
var lookupByVolID = ;
var userIN = [];
var placeIN = [];
var volIN = [];
for (var i = 0; i < results.length; i++)

   var result = results[i];
   if (!lookupByUserID[result.id_user_creator])
       lookupByUserID[result.id_user_creator] = [];
   lookupByUserID[result.id_user_creator].push(result);

   if(!lookupByPlaceID[result.id_place])
      lookupByPlaceID[result.id_place] = [];
   lookupByPlaceID[result.id_place].push(result);

   lookupByVolID[result.id_vol] = result;
   userIN.push(result.id_user_creator);
   placeIN.push(result.id_place);
   volIN.push(result.id_vol);
   results.comments = [];


var userResults = db.queryResults('SELECT users.id_user, users.login, users.verified, users.photo_url FROM users WHERE id_user IN ('+userIN.join(',')+')');
var placeResults = db.queryResults('SELECT place.id_place, place.name AS placeName, place.lat, place.long FROM place WHERE id_place IN ('+placeIN.join(',')+')');
var commentResults = db.queryResults('SELECT comments.id_vol, comments.message, comments.id_user FROM comments WHERE id_vol IN ('+volIN.join(',')+')');
for (var i = 0; i < userResults.length; i++)

   var user = userResults[i];
    for (var j = 0; j < lookupByUserID[user.id_user].length; j++)
        lookupByUserID[user.id_user][j].created_by_user = user;

for (var i = 0; i < placeResults.length; i++)

   var place = placeResults[i];
    for (var j = 0; j < lookupByPlaceID[place.id_place].length; j++)
        lookupByPlaceID[place.id_place][j].place = place;

for (var i = 0; i < commentResults.length; i++)

    var comment = commentResults[i];
    lookupByVolID[comment.id_vol].comments.push(comment);

现在对象“结果”应该包含所有信息