Mongo DB sorting With case insensitive

问题:

I would like to get a sorted form of names from mongodb.I had done this through the following

query.sort().on("name", Order.ASCENDING)

with this query execution I was able to found the sorted results with case sensitive. But i Just want to obtain the results in case ignored form. How to do it? Please guide me through this.

I'm working on java code. So please suggest me with appropriate methods.

回答1:

Update: This answer is out of date, 3.4 will have case insensitive indexes. Look to the JIRA for more information https://jira.mongodb.org/browse/SERVER-90

---

Unfortunately MongoDB does not yet have case insensitive indexes: https://jira.mongodb.org/browse/SERVER-90 and the task has been pushed back.

This means the only way to sort case insensitive currently is to actually create a specific "lower cased" field, copying the value (lower cased of course) of the sort field in question and sorting on that instead.

回答2:

Sorting does work like that in MongoDB but you can do this on the fly with aggregate:

Take the following data:

{ "field" : "BBB" }
{ "field" : "aaa" }
{ "field" : "AAA" }

So with the following statement:

db.collection.aggregate([
    { "$project": {
       "field": 1,
       "insensitive": { "$toLower": "$field" }
    }},
    { "$sort": { "insensitive": 1 } }
])

Would produce results like:

{
    "field" : "aaa",
    "insensitive" : "aaa"
},
{
    "field" : "AAA",
    "insensitive" : "aaa"
},
{
    "field" : "BBB",
    "insensitive" : "bbb"
}

The actual order of insertion would be maintained for any values resulting in the same key when converted.

回答3:

This has been an issue for quite a long time on MongoDB JIRA, but it is solved now. Take a look at this release notes for detailed documentation. You should use collation.

User.find()
    .collation({locale: "en" }) //or whatever collation you want
    .sort({name:'asc'})
    .exec(function(err, users) {
        // use your case insensitive sorted results
    });

回答4:

Here it is in Java. I mixed no-args and first key-val variants of BasicDBObject just for variety

        DBCollection coll = db.getCollection("foo");

        List<DBObject> pipe = new ArrayList<DBObject>();

        DBObject prjflds = new BasicDBObject();
        prjflds.put("field", 1);
        prjflds.put("insensitive", new BasicDBObject("$toLower", "$field"));

        DBObject project = new BasicDBObject();
        project.put("$project", prjflds);
        pipe.add(project);

        DBObject sort = new BasicDBObject();
        sort.put("$sort", new BasicDBObject("insensitive", 1));
        pipe.add(sort);

        AggregationOutput agg = coll.aggregate(pipe);

        for (DBObject result : agg.results()) {
            System.out.println(result);
        }

回答5:

We solve this problem with the help of .sort function in JavaScript array

Here is the code

    function foo() {
      let results = collections.find({
        _id: _id
      }, {
        fields: {
          'username': 1,
        }
      }).fetch();

      results.sort((a, b)=>{
        var nameA = a.username.toUpperCase();
        var nameB = b.username.toUpperCase();

        if (nameA  nameB) {
          return 1;
        }
        return 0;
      });

      return results;
    }