mongo find function mismatch

Question

Mongo version : 3.2.8

My sample json is as below

My query to fetch name equal to apple doesn't work.

db.collection.find( { "products.foods.name": "apple" } )

Instead it fetches all the records, strange?

Neither does the $eq, $lt or $gt work. They result with the entire data.

db.aggregation.find( { "products.foods.min_price": {$eq:10} } )

Thanks in advance.

As others already told you, matching a part of the document returns the complete document unless you take special measures like using the aggregation framework. However, as you seem to regard the food-entries as single records and expect them to be returned individually from that query, my guess is that you have a rather fundamental misunderstanding regarding documents in mongo. The JSON you posted looks like one document to hold it all, which about equals wrtiting all your data in a single row in a relational database. Consider a separate collection for food and individual docs per entry. — mtj, Nov 10 '16 at 13:20

score 1 · Answer 1 · edited May 23 '17 at 12:33

If your entire document is in an _id, then if the query matches db.collection.find( { "products.foods.name": "apple" } ) even though it is a document in foods array the entire document will be displayed, so that you are getting other fruits as well.

To Solve this first use $unwind the aggregation pipeline to break the foods array into individual documents and then use $match.

Please refer this post, It is a similar question and I have answered the steps in detail in that post.

score 0 · Answer 2 · edited May 23 '17 at 12:33

0

Try with this:

db.test.aggregate([
    {$match: {'products.foods.name': 'apple'}}])

Taken from Retrieve only the queried element in an object array in MongoDB collection

You can try other examples from that post.

edited May 23 '17 at 12:33

Community

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answered Nov 10 '16 at 12:32

Kumar Sourav

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score 0 · Answer 3 · answered Nov 10 '16 at 12:51

0

Try with this:db.Exercise.aggregate([ {$match: {'products.foods.min_price': 10}}])

answered Nov 10 '16 at 12:51

Mital Gajjar

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score 0 · Accepted Answer · answered Nov 14 '16 at 09:56

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The solution is to n $unwind both the arrays.

db.foods.aggregate([   
    { $unwind : "$products" },
    { $unwind : "$products.foods" },
    { $match  : { "products.foods.min_price": 10        }}
])

answered Nov 14 '16 at 09:56

Some Java Guy

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mongo find function mismatch

4 Answers4