MongoDB Aggregation: How to get total records coun

问题:

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),
));

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

回答1:

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 ) ) ) )
))

Result will look something like this:

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

回答2:

Use this to find total count in collection.

db.collection.aggregate( [
{ $match : { score : { $gt : 70, $lte : 90 } } },
{ $group: { _id: null, count: { $sum: 1 } } }
] );

回答3:

Since v.3.4 (i think) MongoDB has now a new aggregation pipeline operator named 'facet' which in their own words:

Processes multiple aggregation pipelines within a single stage on the same set of input documents. Each sub-pipeline has its own field in the output document where its results are stored as an array of documents.

In this particular case, this means that one can do something like this:

$result = $collection->aggregate([
  { ...execute queries, group, sort... },
  { ...execute queries, group, sort... },
  { ...execute queries, group, sort... },
  $facet: {
    paginatedResults: [{ $skip: skipPage }, { $limit: perPage }],
    totalCount: [
      {
        $count: 'count'
      }
    ]
  }
]);

The result will be (with for ex 100 total results):

[
  {
    "paginatedResults":[{...},{...},{...}, ...],
    "totalCount":[{"count":100}]
  }
]

回答4:

You can use toArray function and then get its length for total records count.

db.CollectionName.aggregate([....]).toArray().length

回答5:

I did it this way:

db.collection.aggregate([
     { $match : { score : { $gt : 70, $lte : 90 } } },
     { $group: { _id: null, count: { $sum: 1 } } }
] ).map(function(record, index){
        print(index);
 });

The aggregate will return the array so just loop it and get the final index .

And other way of doing it is:

var count = 0 ;
db.collection.aggregate([
{ $match : { score : { $gt : 70, $lte : 90 } } },
{ $group: { _id: null, count: { $sum: 1 } } }
] ).map(function(record, index){
        count++
 }); 
print(count);

回答6:

Solution provided by @Divergent does work, but in my experience it is better to have 2 queries:

1. First for filtering and then grouping by ID to get number of filtered elements. Do not filter here, it is unnecessary.
2. Second query which filters, sorts and paginates.

Solution with pushing $$ROOT and using $slice runs into document memory limitation of 16MB for large collections. Also, for large collections two queries together seem to run faster than the one with $$ROOT pushing. You can run them in parallel as well, so you are limited only by the slower of the two queries (probably the one which sorts).

I have settled with this solution using 2 queries and aggregation framework (note - I use node.js in this example, but idea is the same):

var aggregation = [
  {
    // If you can match fields at the begining, match as many as early as possible.
    $match: {...}
  },
  {
    // Projection.
    $project: {...}
  },
  {
    // Some things you can match only after projection or grouping, so do it now.
    $match: {...}
  }
];


// Copy filtering elements from the pipeline - this is the same for both counting number of fileter elements and for pagination queries.
var aggregationPaginated = aggregation.slice(0);

// Count filtered elements.
aggregation.push(
  {
    $group: {
      _id: null,
      count: { $sum: 1 }
    }
  }
);

// Sort in pagination query.
aggregationPaginated.push(
  {
    $sort: sorting
  }
);

// Paginate.
aggregationPaginated.push(
  {
    $limit: skip + length
  },
  {
    $skip: skip
  }
);

// I use mongoose.

// Get total count.
model.count(function(errCount, totalCount) {
  // Count filtered.
  model.aggregate(aggregation)
  .allowDiskUse(true)
  .exec(
  function(errFind, documents) {
    if (errFind) {
      // Errors.
      res.status(503);
      return res.json({
        'success': false,
        'response': 'err_counting'
      });
    }
    else {
      // Number of filtered elements.
      var numFiltered = documents[0].count;

      // Filter, sort and pagiante.
      model.request.aggregate(aggregationPaginated)
      .allowDiskUse(true)
      .exec(
        function(errFindP, documentsP) {
          if (errFindP) {
            // Errors.
            res.status(503);
            return res.json({
              'success': false,
              'response': 'err_pagination'
            });
          }
          else {
            return res.json({
              'success': true,
              'recordsTotal': totalCount,
              'recordsFiltered': numFiltered,
              'response': documentsP
            });
          }
      });
    }
  });
});

回答7:

Use $count operator to get the total count of aggregation result

Query :

db.collection.aggregate(
  [
    {
      $match: {
        ...
      }
    },
    {
      $aggregate: {
        ...
      }
    },
    {
      $count: "totalCount"
    }
  ]
)

Result:

{
   "totalCount" : Number of records (some integer value)
}

回答8:

Sorry, but I think you need two queries. One for total views and another one for grouped records.

You can find useful this answer