My table name is : table_video My field name in db is : url_value

Whose value is : http://192.168.1.124/test/abcd/abcd.m3u8

Value which is needed by me : http://192.168.1.124/test/abcd_NEW/abcd_NEW.m3u8

There are multiple values in place of "abcd", above url is just one example.

var cursor = db.table_video.find();
while (cursor.hasNext()) {
  var x = cursor.next();
  print("\n\n-----------------------------------");

  print("Before : url_value : "+x['url_value']);

  x['url_value'] = x['url_value'].replace(/^(.*?)\/test\/(.*?)\/(.*?)\.m3u8$/g, $1/test/$2_NEW/$2_NEW.m3u8);

  print("After : url_value : "+x['url_value']);

  db.table_video.update({_id : x._id}, x);

}

When I execute above mand in mongo console, it gives an error : 2015-11-28T12:40:08.342+0530 ReferenceError: $1 is not defined

Any help is greatly appreciated

My table name is : table_video My field name in db is : url_value

Whose value is : http://192.168.1.124/test/abcd/abcd.m3u8

Value which is needed by me : http://192.168.1.124/test/abcd_NEW/abcd_NEW.m3u8

There are multiple values in place of "abcd", above url is just one example.

var cursor = db.table_video.find();
while (cursor.hasNext()) {
  var x = cursor.next();
  print("\n\n-----------------------------------");

  print("Before : url_value : "+x['url_value']);

  x['url_value'] = x['url_value'].replace(/^(.*?)\/test\/(.*?)\/(.*?)\.m3u8$/g, $1/test/$2_NEW/$2_NEW.m3u8);

  print("After : url_value : "+x['url_value']);

  db.table_video.update({_id : x._id}, x);

}

When I execute above mand in mongo console, it gives an error : 2015-11-28T12:40:08.342+0530 ReferenceError: $1 is not defined

Any help is greatly appreciated

• javascript
• regex
• mongodb
• replace
• mongodb-query

Share Improve this question Follow edited Nov 28, 2015 at 8:24 Sede 61.3k2020 gold badges158158 silver badges161161 bronze badges asked Nov 28, 2015 at 6:43 gehlotpareshgehlotparesh 14233 silver badges1414 bronze badges 6

• 2 Check This – Narendrasingh Sisodia Commented Nov 28, 2015 at 6:49
• My regex is already working in Kate, even I checked the link you provided, however it's giving error in mongodb – gehlotparesh Commented Nov 28, 2015 at 6:55
• @user2767817 your regex is wrong it will not work in Mongo. use the one provided by Uchiha. – Sede Commented Nov 28, 2015 at 7:03
• Regex provided by Uchiha is right, however, when I use in mongodb, it gives error : 2015-11-28T12:40:08.342+0530 ReferenceError: $1 is not defined – gehlotparesh Commented Nov 28, 2015 at 7:15
• please edit your question to include the new code – Sede Commented Nov 28, 2015 at 7:19

 |  Show 1 more ment

2 Answers 2

Sorted by: Reset to default 4

Your regex is wrong you can get the expected result using this regex¹.

But the best way to do this is using "bulk" operations for maximum efficiency.

var bulk = db.table_video.initializeUnorderedBulkOp();
var count = 0;
db.table_video.find().forEach(function(doc) {
    var newUrlValue = doc.url_value.replace(/^(.*?)\/test\/(.*?)\/(.*?)\.m3u8$/, '$1/test/$2_NEW/$3_NEW.m3u8');
    bulk.find( { '_id': doc._id } ).updateOne( { 
        '$set': { 'url_value': newUrlValue } 
    });
    count++;
    if (count % 100 === 0) {
        // Execute per 100 operation and re-init
        bulk.execute();
        bulk = db.table_video.initializeUnorderedBulkOp();
        count = 0;
    }
})

// Clean up queues
if (count > 0)  bulk.execute();

---

If your MongoDB version is older that 2.6 you need to use a while loop.

var cursor = db.table_video.find();
while (cursor.hasNext()) {
    var x = cursor.next();
    print("\n\n-----------------------------------");
    print("Before : url_value : "+x['url_value']);
    var newUrlValue = x['url_value'].replace(/^(.*?)\/test\/(.*?)\/(.*?)\.m3u8$/, '$1/test/$2_NEW/$2_NEW.m3u8');
    print("After : url_value : "+newUrlValue);
    db.table_video.update({ _id : x._id }, { '$set': { 'url_value': newUrlValue } } );

}

---

1. Regex provided by @Uchiha.

in mongodb version 4.2 you have regexFind project operator which can be used to get the matches, then you ocan use substr to replace parts of the pattern