Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
399 views
in Technique[技术] by (71.8m points)

javascript - MongoDB group by hour

I save tweets to mongo DB:

 twit.stream('statuses/filter', {'track': ['animal']}, function(stream) {
    stream.on('data', function(data) {
        console.log(util.inspect(data));

        data.created_at = new Date(data.created_at);
        collectionAnimal.insert(data, function(err, docs) {});
    });
});

It's OK.

The tweet time in MongoDB is in format: 2014-04-25 11:45:14 GMT (column created_at) Now I need group column created_at in hours. I would like to have the result:

hour | count tweets in hour


1 | 28

2 | 26

3 | 32

4 | 42

5 | 36

...

My unsuccessful attempt:

    $keys = array('created_at' => true);
    $initial = array('count' => 0);
    $reduce = "function(doc, prev) { prev.count += 1 }";

    $tweetsGroup = $this->collectionAnimal->group( $keys, $initial, $reduce );

But my not able to group by hour.

How to do it?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

I could tell you how you can group using aggregation framework directly on mongo console

db.tweets.aggregate(
 { "$project": {
      "y":{"$year":"$created_at"},
      "m":{"$month":"$created_at"},
      "d":{"$dayOfMonth":"$created_at"},
      "h":{"$hour":"$created_at"},
      "tweet":1 }
 },
 { "$group":{ 
       "_id": { "year":"$y","month":"$m","day":"$d","hour":"$h"},
       "total":{ "$sum": "$tweet"}
   }
 })

For more options you can look here: http://docs.mongodb.org/manual/reference/operator/aggregation-date/

You will also need to find appropriate way of of using aggregation framework from whichever programming language you are using.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...