如何使用MongoDB和PHP聚合将数组与嵌套数组分开？

问题描述

我想获取不同的code_y值，并且我使用下面的代码，输出等于select distinct code_y from table_x，

$res = $db_x->command (
        array(
            "aggregate" => "table_x","pipeline" =>
                array(
                    array( '$group' => array( "_id" => ['id' =>'$code_y','name' => '$code_x','color' => '$color']))),"cursor" => ['batchSize' => 200]
        )
);

我得到的结果如下，我不需要[_id] => Array，只需ID，名称和颜色，如何使用MongoDB和PHP聚合将数组与嵌套数组分开？或者如何将[_id] => Array与id,name and color放在一起？就像在我的前端htmls页面中一样，由于这些嵌套数组，我得到了[object] [Object]。

---

   [0] => Array
        (
            [_id] => Array
                (
                    [id] => a1
                    [name] => bbb
                    [color] => blue
                )

        )

    [1] => Array
        (
            [_id] => Array
                (
                    [id] => a2
                    [name] => aaa
                    [color] => blue
                )

        )

    [2] => Array
        (
            [_id] => Array
                (
                    [id] => a3
                    [name] => abc
                    [color] =>red
                )

        )

---

我想要的如下：

[0] => Array
    (
       [id] => a1
       [name] => bbb
       [color] => blue
    )

[1] => Array
    (
       [id] => a2
       [name] => aaa
       [color] => blue


    )

[2] => Array
    (
      [id] => a3
      [name] => abc
      [color] =>red
    )

解决方法

您需要的是小组之后的又一个阶段。

您可以使用项目并列出每个字段：

array( '$project' => array( 
     "id" => '$_id.id',"name" => '$_id.name',"color" => '$_id.color'
))

或$replaceRoot来抓住他们：

array( '$replaceRoot' => array( "newRoot" => '$_id' ))