How to query MongoDB with "like"?

Freewind Source

I want to query something as SQL's like query:

select * 
from users 
where name like '%m%'

How to do the same in MongoDB?
I can't find a operator for like in the documentations.



answered 8 years ago Joshua Partogi #1

You would use regex for that in mongo.

e.g: db.users.find({"name": /^m/})

answered 8 years ago Kyle H #2

That would have to be:

db.users.find({"name": /.*m.*/})

or, similar:

db.users.find({"name": /m/})

You're looking for something that contains "m" somewhere (SQL's '%' operator is equivalent to Regexp's '.*'), not something that has "m" anchored to the beginning of the string.

answered 7 years ago leon #3

In PHP, you could use following code:

$collection->find(array('name'=> array('$regex' => 'm'));

answered 6 years ago Afshin Mehrabani #4


  • PyMongo using Python
  • Mongoose using Node.js
  • Jongo, using Java
  • mgo, using Go

you can do:

db.users.find({'name': {'$regex': 'sometext'}})

answered 5 years ago Crasher #5

You can use where statement to build any JS script:

db.myCollection.find( { $where: "'m') >= 0" } );


answered 4 years ago Eddy #6

If using node.js, it says that you can write this:

db.collection.find( { field: /acme.*corp/i } );
db.collection.find( { field: { $regex: 'acme.*corp', $options: 'i' } } );

Also, you can write this:

db.collection.find( { field: new RegExp('acme.*corp', 'i') } );

answered 4 years ago user2312578 #7

In Go and the mgo driver:

Collection.Find(bson.M{"name": bson.RegEx{"m", ""}}).All(&result)

where result is the struct instance of the sought after type

answered 4 years ago MAA #8

In SQL, the ‘like’ query is looks like this :

select * from users where name like '%m%'

In MongoDB console, it looks like this :

db.users.find({"name": /m/})     // Not JSON formatted

db.users.find({"name": /m/}).pretty()  // JSON formatted

In addion pretty() method will in all the places where produce formatted JSON structure which is more readable.

answered 4 years ago Johnathan Douglas #9

db.users.insert({name: 'paulo'})
db.users.insert({name: 'patric'})
db.users.insert({name: 'pedro'})

db.users.find({name: /a/})  //like '%a%'

out: paulo, patric

db.users.find({name: /^pa/}) //like 'pa%' 

out: paulo, patric

db.users.find({name: /ro$/}) //like '%ro'

out: pedro

answered 4 years ago cmarrero01 #10

You can use the new feature of 2.6 mongodb:{desc: "This is a string with text"});{desc:"This is a another string with Text"});{"desc":"text"});{

answered 4 years ago Dap #11

For PHP mongo Like.
I had several issues with php mongo like. i found that concatenating the regex params helps in some situations PHP mongo find field starts with. I figured I would post on here to contribute to the more popular thread


db()->users->insert(['name' => 'john']);
db()->users->insert(['name' => 'joe']);
db()->users->insert(['name' => 'jason']);

// starts with
$like_var = 'jo';
$prefix = '/^';
$suffix = '/';
$name = $prefix . $like_var . $suffix;
db()->users->find(['name' => array('$regex'=>new MongoRegex($name))]);
output: (joe, john)

// contains
$like_var = 'j';
$prefix = '/';
$suffix = '/';
$name = $prefix . $like_var . $suffix;
db()->users->find(['name' => array('$regex'=>new MongoRegex($name))]);

output: (joe, john, jason)

answered 4 years ago The6thSense #12

Already u got the answers but to match regex with case insensitivity

You could use the following query

db.users.find ({ "name" : /m/i } ).pretty()

The i in the /m/i indicates case insensitivity and .pretty() provides a more pretty output

answered 3 years ago Vaibhav #13

If you are using Spring-Data Mongodb You can do this in this way:

String tagName = "m";
Query query = new Query();

answered 3 years ago prayagupd #14

Like Query would be as shown below

db.movies.find({title: /.*Twelve Monkeys.*/}).sort({regularizedCorRelation : 1}).limit(10);

for scala ReactiveMongo api,

val query = BSONDocument("title" -> BSONRegex(".*"+name+".*", "")) //like
val sortQ = BSONDocument("regularizedCorRelation" -> BSONInteger(1))
val cursor = collection.find(query).sort(sortQ).options(QueryOpts().batchSize(10)).cursor[BSONDocument]

answered 3 years ago Shaishab Roy #15

In nodejs project and use mongoose use Like query

var User = mongoose.model('User');

var searchQuery={}; =; = {$regex:, $options: 'i'};
User.find(searchQuery, function(error, user) {
                if(error || user === null) {
                    return res.status(500).send(error);
                return res.status(200).send(user);

answered 3 years ago Aqib Mumtaz #16

For Mongoose in Node.js

db.users.find({'name': {'$regex': '.*sometext.*'}})

answered 2 years ago Shalabh Raizada #17

Use regular expressions matching as below. The 'i' shows case insensitivity.

var collections = mongoDatabase.GetCollection("Abcd");

var queryA = Query.And(
         Query.Matches("strName", new BsonRegularExpression("ABCD", "i")), 
         Query.Matches("strVal", new BsonRegularExpression("4121", "i")));

var queryB = Query.Or(
       Query.Matches("strName", new BsonRegularExpression("ABCD","i")),
       Query.Matches("strVal", new BsonRegularExpression("33156", "i")));

var getA = collections.Find(queryA);
var getB = collections.Find(queryB);

answered 2 years ago sravanthi #18

As Mongo shell support regex, that's completely possible.

db.users.findOne({"name" : /.*sometext.*/});

If we want the query to be case-insensitive, we can use "i" option, like shown below:

db.users.findOne({"name" : /.*sometext.*/i});

answered 2 years ago Somnath Muluk #19

There are already many answers. I am giving different types of requirements and solutions for string search with regex.

You can do with regex which contain word i.e like. Also you can use $options => i for case insensitive search

Contains string

db.collection.find({name:{'$regex' : 'string', '$options' : 'i'}})

Doesn't Contains string only with regex

db.collection.find({name:{'$regex' : '^((?!string).)*$', '$options' : 'i'}})

Exact case insensitive string

db.collection.find({name:{'$regex' : '^string$', '$options' : 'i'}})

Start with string

db.collection.find({name:{'$regex' : '^string', '$options' : 'i'}})

End with string

db.collection.find({name:{'$regex' : 'string$', '$options' : 'i'}})

Keep this as a bookmark, and a reference for any other alterations you may need.

answered 2 years ago jarry jafery #20

If you want 'Like' search in mongo then you should go with $regex by using this query will be


for more you can read the documentation as well.

answered 1 year ago Lakmal Vithanage #21

I found a free tool to translate MYSQL queries to MongoDB. I checked with several queries. as i see almost all them are correct. According to that, The answer is

    "name": "%m%"

answered 1 year ago Bruno Bronosky #22

It seems that there are reasons for using both the javascript /regex_pattern/ pattern as well as the mongo {'$regex': 'regex_pattern'} pattern. See: MongoBD RegEx Syntax Restrictions

This is not a complete RegEx tutorial, but I was inspired to run these tests after seeing a highly voted ambiguous post above.

> ['abbbb','bbabb','bbbba'].forEach(function(v){db.test_collection.insert({val: v})})

> db.test_collection.find({val: /a/})
{ "val" : "abbbb" }
{ "val" : "bbabb" }
{ "val" : "bbbba" }

> db.test_collection.find({val: /.*a.*/})
{ "val" : "abbbb" }
{ "val" : "bbabb" }
{ "val" : "bbbba" }

> db.test_collection.find({val: /.+a.+/})
{ "val" : "bbabb" }

> db.test_collection.find({val: /^a/})
{ "val" : "abbbb" }

> db.test_collection.find({val: /a$/})
{ "val" : "bbbba" }

> db.test_collection.find({val: {'$regex': 'a$'}})
{ "val" : "bbbba" }

answered 1 year ago Albert s #23

MongoRegex has been deprecated.
Use MongoDB\BSON\Regex

$regex = new MongoDB\BSON\Regex ( '^m');
$cursor = $collection->find(array('users' => $regex));
//iterate through the cursor

answered 1 year ago CEDA #24

If you're using PHP, you can use MongoDB_DataObject wrapper like below:

$model = new MongoDB_DataObject();

$model->query("select * from users where name like '%m%'");

while($model->fetch()) {


$model = new MongoDB_DataObject('users);

$model->whereAdd("name like '%m%'");


while($model->fetch()) {

answered 1 year ago damd #25

With MongoDB Compass, you need to use the strict mode syntax, as such:

{ "text": { "$regex": "^Foo.*", "$options": "i" } }

(In MongoDB Compass, it's important that you use " instead of ')

answered 1 year ago user3645907 #26

You have 2 choices:

db.users.find({"name": /string/})


db.users.find({"name": {"$regex": "string", "$options": "i"}})

On second one you have more options, like "i" in options to find using case insensitive. And about the "string", you can use like ".string." (%string%), or "string.*" (string%) and ".*string) (%string) for example. You can use regular expression as you want.


answered 1 year ago Shubham Verma #27

FullName like 'last' with status==’Pending’ between two dates:


status== 'Pending' and orderId LIKE ‘PHA876174’:


answered 12 months ago priya raj #28

db.customer.find({"customerid": {"$regex": "CU_00000*", "$options": "i"}}).pretty()

When we are searching for string patterns, always it is better to use the above pattern as when we are not sure about case. Hope that helps!!!

answered 7 months ago dinesh vishe #29

>'name') [ "honda", "tat", "tata", "tata3" ]

>{"name":/. ta. /})

answered 5 months ago kz_sergey #30

Use aggregation substring search (with index!!!):

        $project : {
            fieldExists : {
                $indexOfBytes : ['$field', 'string']
    }, {
        $match : {
            fieldExists : {
                $gt : -1
    }, {
        $limit : 5

answered 1 week ago user3284463 #31

Regex are expensive are process.

Another way is to create an index of text and then search it using $search.

Create a text Index of fields you want to make searchable:

db.collection.createIndex({name: 'text', otherField: 'text'});

Search for a string in text index:

  '$text'=>{'$search': "The string"}

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