CODEWITHSUNDEEP
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ArminMz
ArminMz

Reputation: 345

Can I use $in statement for two fields instead of one in mongo db?

I had a list of dictionaries that had one field called unique_field_id and I wanted to get each document for each item of this list. One way was to do it with a for loop and calling find_one() query like this:

my_list = [{'unique_field_id': 1}, {'unique_field_id': 2}, {'unique_field_id': 3}]
document_list = list()

for el in my_list:
    document = db.collection_name.find_one(el)
    document_list.append(document)

The other way and the better and faster way was to do it like this:

my_list = [{'unique_field_id': 1}, {'unique_field_id': 2}, {'unique_field_id': 3}]
my_list_values = [d['unique_field_id'] for d in my_list]
document_list = db.collection_name.find({'unique_field_id': {'$in': my_list_values}})

This was working fine with me until unique field of my documents changed to two fields. Now I have a list of dictionaries with two fields unique_field_id and other_unique_field_id which are indexed together as the unique field of my document.

If I want to take the first approach it would be like this:

my_list = [{'unique_field_id': 1, 'other_unique_field_id': 'a'}, {'unique_field_id': 2, 'other_unique_field_id': 'b'}, {'unique_field_id': 3, 'other_unique_field_id': 'c'}]
document_list = list()
for el in my_list:
    document = db.collection_name.find_one(el)
    document_list.append(document)

BUT I need to do it the second way and check the $in statement for index of two fields instead of doing it one by one because it is much faster. How Can I Do It? Also I think that there is useful answer in this question which I don't understand fully. Mongo $in with compound index

Upvotes: 0

Views: 165

Answers (1)

hhharsha36
hhharsha36

Reputation: 3349

Use the $or operator to find the command.

my_list = [
    {'unique_field_id': 1, 'other_unique_field_id': 'a'},
    {'unique_field_id': 2, 'other_unique_field_id': 'b'},
    {'unique_field_id': 3, 'other_unique_field_id': 'c'}
]

document_list = list(db.collection_name.find({
    "$or": my_list
}))

Upvotes: 1

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