mongodb find all document in which one field contains another field

Asked Jan 25 '19 at 09:18

Active Jan 25 '19 at 10:27

Viewed 547 times

I have documents in mongodb like:

{
  "legal_name": "Roshan Bhandari",
  "first_name": "Roshan"
}

In sql for such data I would do:

Select * from mytable where first_name not like '%' + LOWER([legal_name]) + '%'

In mongodb I tried following:

db.mytable.find({'$expr': {'first_name':{'$nin': 'legal_name'}})

But it didnt work. What would be approriate solution for this?

edited Jan 25 '19 at 10:27

asked Jan 25 '19 at 09:18

bor

2

`legal_name` should be in quotes – zmbq Jan 25 '19 at 09:20
@zmbq that is not the case, please refer to sql I wrote. – bor Jan 25 '19 at 10:41
@VaghaniJanak I do not want string comparison, I want 2 column comparison – bor Jan 25 '19 at 10:41
db.mytable.find({ $where : "this.first_name != this.legal_name" }) – Vaghani Janak Jan 25 '19 at 10:52
1

Please refers below link https://stackoverflow.com/questions/4442453/mongodb-query-condition-on-comparing-2-fields – Vaghani Janak Jan 25 '19 at 10:55
You can do something like this https://mongoplayground.net/p/cOxOZNLvRRx – Ashh Jan 25 '19 at 19:06

0 Answers0