Javascript 将Sql请求转换为json，但使用特定格式

我的问题是：我可以有一个Json格式，但不是我想要的格式。我向您展示我的代码，然后向您展示我想要的Json模型。我不知道如何继续，如果我必须更改php代码、js代码或数据库，我认为fetchAll（）数组不适合我。我真的需要帮助。。。代码正在运行，只是格式不好。我只对sql请求和数据库使用AngularJs、Html、Css和php。我告诉你我在使用什么，因为它可能存在Js或php中的解决方案，甚至是我的sql请求，这可能是不合适的

我的php代码：

$request = $bdd->prepare('
    SELECT z.name zoneName, c.name columnName, w.name woodName 
    FROM zones z
    JOIN columns c ON (c.ID_zone = z.ID)
    JOIN woods w ON (w.ID_column = c.ID)
    ');

$request->execute();
$result = $request->fetchAll(PDO::FETCH_ASSOC);
echo json_encode($result);

我在Json中的结果：

        [{"zoneName":"A1","columnName":"C1","woodName":"10-27-050-03\/17"},
        {"zoneName":"A1","columnName":"C1","woodName":"42-27-050-03\/17"},
        {"zoneName":"A1","columnName":"C2","woodName":"20-23-050-03\/17"},
        {"zoneName":"A2","columnName":"C3","woodName":"90-37-050-03\/17"}]

我想要的结果是：

{"zoneName":[{"columnName": [[{"id":"woodName"},{"id":"woodName"},...]]},{"columnName": [[{"id":"woodName"},{"id":"woodName"}, ...]]},...],"zoneName":[{"columnName": [[{"id":"woodName"},{"id":"woodName"}, ...]]},{"columnName": [[{"id":"woodName"},{"id":"woodName"}, ...]]},...],...}

---

我终于成功了，但不是以我想要的方式。。。然后我不想解决这个问题，因为我直接从我的请求中创建了自己的Json。我仍然想要一个解决方案的方式是使用sql请求，该请求将通过php函数fetchAll（）+Json_encode（）直接返回我想要的Json

我所做的代码（有效）：

我知道我的代码不干净，但我没有足够的编程经验

$outp的结果：

{"A1":[{"type":"container","id":"C1","columns":[[{"type":"item","id":"10-27-050-03_17"},{"type":"item","id":"10-28-050-03_17"}]]},{"type":"container","id":"C2","columns":[[{"type":"item","id":"20-23-050-03_17"},{"type":"item","id":"20-23-050-03_18"}]]}],"A2":[{"type":"container","id":"C1","columns":[[{"type":"item","id":"30-27-052-03_17"},{"type":"item","id":"30-27-052-03_17"}]]},{"type":"container","id":"C2","columns":[[{"type":"item","id":"40-27-050-03_17"},{"type":"item","id":"40-28-050-03_17"}]]},{"type":"container","id":"C3","columns":[[{"type":"item","id":"50-27-050-03_17"},{"type":"item","id":"50-28-050-03_17"}]]}],"A3":[{"type":"container","id":"C1","columns":[[{"type":"item","id":"60-27-050-03_17"},{"type":"item","id":"60-27-050-03_17"}]]},{"type":"container","id":"C2","columns":[[{"type":"item","id":"70-27-050-03_17"},{"type":"item","id":"70-27-050-03_17"}]]}],"A4":[{"type":"container","id":"C1","columns":[[{"type":"item","id":"80-27-050-03_17"},{"type":"item","id":"80-28-050-03_17"}]]},{"type":"container","id":"C2","columns":[[{"type":"item","id":"90-27-040-03_17"},{"type":"item","id":"90-27-050-03_17"}]]}]}

即使我得到了我想要的输出，我仍然在寻找解决方案，因为现在操作我的数据库将非常困难

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祝你有愉快的一天

[“id”：“woodName”，“id”：“woodName”，…]

不可能。数组/对象不包含重复的键，您永远不会以这种方式获得json响应。你不喜欢你得到的结果是什么？当然我犯了一个错误，谢谢B.Desai，我编辑了它。我不喜欢我得到的结果，因为我正在创建一个应用程序，其中有一块zoneName，它包含了它的所有columnName，它包含了它的所有woodName，所以有了这个json，我不能按照我的意愿嵌套它，直到等待解决方案，有人吗？

{"A1":[{"type":"container","id":"C1","columns":[[{"type":"item","id":"10-27-050-03_17"},{"type":"item","id":"10-28-050-03_17"}]]},{"type":"container","id":"C2","columns":[[{"type":"item","id":"20-23-050-03_17"},{"type":"item","id":"20-23-050-03_18"}]]}],"A2":[{"type":"container","id":"C1","columns":[[{"type":"item","id":"30-27-052-03_17"},{"type":"item","id":"30-27-052-03_17"}]]},{"type":"container","id":"C2","columns":[[{"type":"item","id":"40-27-050-03_17"},{"type":"item","id":"40-28-050-03_17"}]]},{"type":"container","id":"C3","columns":[[{"type":"item","id":"50-27-050-03_17"},{"type":"item","id":"50-28-050-03_17"}]]}],"A3":[{"type":"container","id":"C1","columns":[[{"type":"item","id":"60-27-050-03_17"},{"type":"item","id":"60-27-050-03_17"}]]},{"type":"container","id":"C2","columns":[[{"type":"item","id":"70-27-050-03_17"},{"type":"item","id":"70-27-050-03_17"}]]}],"A4":[{"type":"container","id":"C1","columns":[[{"type":"item","id":"80-27-050-03_17"},{"type":"item","id":"80-28-050-03_17"}]]},{"type":"container","id":"C2","columns":[[{"type":"item","id":"90-27-040-03_17"},{"type":"item","id":"90-27-050-03_17"}]]}]}