MongoDB聚合:如何获取总记录数? [英] MongoDB Aggregation: How to get total records count?

问题描述

我已经使用聚合从mongodb中获取记录.

I have used aggregation for fetching records from mongodb.

$result = $collection->aggregate(array(
  array('$match' => $document),
  array('$group' => array('_id' => '$book_id', 'date' => array('$max' => '$book_viewed'),  'views' => array('$sum' => 1))),
  array('$sort' => $sort),
  array('$skip' => $skip),
  array('$limit' => $limit),
));

如果我无限制地执行此查询，则将提取10条记录.但是我想将限制保持为2.因此，我想获取总记录数.如何进行汇总?请给我建议.谢谢

If I execute this query without limit then 10 records will be fetched. But I want to keep limit as 2. So I would like to get the total records count. How can I do with aggregation? Please advice me. Thanks

推荐答案

这是在单个查询中同时获取分页结果和结果总数的最常见问题之一.我无法解释当我终于实现大声笑时的感受.

This is one of the most commonly asked question to obtain the paginated result and the total number of results simultaneously in single query. I can't explain how I felt when I finally achieved it LOL.

$result = $collection->aggregate(array(
  array('$match' => $document),
  array('$group' => array('_id' => '$book_id', 'date' => array('$max' => '$book_viewed'),  'views' => array('$sum' => 1))),
  array('$sort' => $sort),

// get total, AND preserve the results
  array('$group' => array('_id' => null, 'total' => array( '$sum' => 1 ), 'results' => array( '$push' => '$$ROOT' ) ),
// apply limit and offset
  array('$project' => array( 'total' => 1, 'results' => array( '$slice' => array( '$results', $skip, $length ) ) ) )
))

结果将如下所示:

[
  {
    "_id": null,
    "total": ...,
    "results": [
      {...},
      {...},
      {...},
    ]
  }
]

这篇关于MongoDB聚合:如何获取总记录数?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！