取“列"的总和在 MongoDB 中

Taking sum of quot;columnquot; in MongoDB(取“列的总和在 MongoDB 中)

问题描述

在 MySQL 中，我只需执行SELECT sum(pts) FROM table"即可获取表中 pts 列的总和.但是，我正在将此应用程序转换为 MongoDB，并且找不到一种直接的方法来获取我的集合中的公共键类型"的总和.我正在使用 PHP，所以请提供 PHP 代码来完成这项工作.

In MySQL, I would simply go "SELECT sum(pts) FROM table" to get the sum of the pts column on the table. However, I am converting this application to MongoDB and cannot find a straightforward way to get a sum of the common key "type" in my collection. I am using PHP, so please give PHP code to make this work.

这是我的数据:

{ "_id" : ObjectId("4f136dab5056f65b61000000"), "p_id" : "3", "g_id" : "1", "type" : "3" }
{ "_id" : ObjectId("4f136dad5056f65760000000"), "p_id" : "8", "g_id" : "1", "type" : "7" }
{ "_id" : ObjectId("4f136daf5056f65860000000"), "p_id" : "6", "g_id" : "1", "type" : "4" }
{ "_id" : ObjectId("4f136db15056f65460000000"), "p_id" : "27", "g_id" : "1", "type" : "2" }

如何让 $sum 等于type"键的总和?

How can I get $sum to equal the total of the "type" key?

推荐答案

利用 MongoDB 新的 聚合框架:

Utilizing MongoDB's new Aggregation Framework:

$result = $db->command(array(

    'aggregate' => 'COLLECTION_NAME',

    'pipeline'  => array(
        // Match:
        // We can skip this in case that we don't need to filter 
        // documents in order to perform the aggregation
        array('$match' => array(
            // Criteria to match against
        )),

        // Group:
        array('$group'  => array(
            // Groupby fields:
            '_id'       => "$fieldname1, $fieldname2, $or_empty, $etc",
            // Count:
            'count'     => array('$sum' => 1),
            // Sum:
            'type_sum'  => array('$sum' => '$type'),
        ))

        // Other aggregations may go here as it's really a PIPE :p

    ),
));

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