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How can I index a collection of records by one of their field?


By : Michael Baker
Date : October 16 2020, 08:10 AM
hope this fix your issue If you want a container where the key itself is part of the value, you can use HashSet with a newtype to reimplement the required traits: Hash, Borrow, PartialEq and Eq.
For example (link to playground):
code :
struct HRecord(Record);
impl Hash for HRecord {
    fn hash<H: Hasher>(&self, state: &mut H) {
        self.0.get_key().hash(state);
    }
}
impl PartialEq for HRecord {
    fn eq(&self, r: &HRecord) -> bool {
        self.0.get_key().eq(r.0.get_key())
    }
}
impl Eq for HRecord {}
impl Borrow<str> for HRecord {
    fn borrow(&self) -> &str {
        self.0.get_key()
    }
}
let mut h = HashSet::new();
h.insert(HRecord(Record { ... }));
let obj = h.get("...").map(|h| &h.0);


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When retrieving records from a MongoDb collection, do the records come back sorted based on the index that was set up fo


By : Manish Jain
Date : March 29 2020, 07:55 AM
I wish did fix the issue. Just as with many other dictionary implementations, order in database is not defined or guaranteed (except for capped collections) and subject to change at any time:

mongodb, select the records with a field that appears at least N times in the collection


By : Артём Гринн
Date : March 29 2020, 07:55 AM
it helps some times If you are on MongoDB v2.2+ I recommend using the new aggregation framework.
Here is the command to extract user.id appearing more than 20 times on the collection.
code :
db.data.aggregate([{$group: {_id: '$user.id', hits: {$sum: 1}}}, {$match: {hits: {$gt: 20}}}])

Drupal Profile2 / Field collection sorting records


By : emiliano bb
Date : March 29 2020, 07:55 AM
This might help you The image below shows a field collection I have made with 3 fields. One the left of each collection item is a drag handle. You can re-order content using these handles.

Get a list of records from a collection sorted by count and uniqueness of a field


By : Michael
Date : March 29 2020, 07:55 AM
should help you out Use the aggregation tools for this. You haven't listed a language in your question, so here's the mongodb command directly. This assumes your collection is named 'urls'.
code :
db.urls.aggregate([{$group: {_id:'$reference', total:{$sum:1} } }]);

How to add a field in all records of mongo db collection?


By : Huajun Bai
Date : March 29 2020, 07:55 AM
seems to work fine Use the $concat operator in the $project step of an aggregation framework pipeline. Running the following aggregation will give you the desired output without need to update your collection:
code :
db.collection.aggregate([
    {
        "$project": {
            "data1": 1,
            "data2": { $concat: [ "$data1", " ", "$data2" ] }
        }
    }
])
var cursor = db.collection.aggregate([
        {
            "$project": {
                "data1": 1,
                "data2": { $concat: [ "$data1", " ", "$data2" ] }
            }
        }
    ]),
    updateCollUsingAgg = function(doc){
        db.collection.update(
            { "_id": doc._id },
            { "$set": { "data2": doc.data2 } }
        )
    }

cursor.forEach(updateCollUsingAgg);
var cursor = db.collection.find(),
    updateCollUsingFind = function(doc){
        db.collection.update(
            { "_id": doc._id },
            { "$set": { "data2": doc.data1+" "+doc.data2 } }
        )
    };
cursor.forEach(updateCollUsingFind);
// Bulk update collection
var bulkUpdateOp = db.collection.initializeUnorderedBulkOp(), 
    pipeline = [
        {
            "$project": {
                "data1": 1,
                "data2": { $concat: [ "$data1", " ", "$data2" ] }
            }
        }
    ],
    counter =  0, // counter to keep track of the batch update size
    // Get modified data2 fields using aggregation framework
    cursor = db.collection.aggregate(pipeline); 

cursor.forEach(function(doc){
    // update collection
    bulkUpdateOp.find({"_id": doc._id}).updateOne({ "$set": { "data2": doc.data2 } }); 
    counter++; // increment counter
    // execute the bulk update operation in batches of 500
    if (counter % 500 == 0) { 
        bulkUpdateOp.execute();
        bulkUpdateOp = db.collection.initializeUnorderedBulkOp();
    }
});

if (counter % 500 != 0) { bulkUpdateOp.execute(); }
 var pipeline = [
        {
            "$project": {
                "data1": 1,
                "data2": { $concat: [ "$data1", " ", "$data2" ] }
            }
        }
    ],
    cursor = db.collection.aggregate(pipeline),
    bulkUpdateOps = cursor.map(function (doc) { 
        return { 
            "updateOne": {
                "filter": { "_id": doc._id },
                "update": { "$set": { "data2": doc.data2 } } 
             }
        };
    });         

db.collection.bulkWrite(bulkUpdateOps);
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