Zulway - 7 months ago 45

Node.js Question

I have a Collection in my database where most documents have an array-field. These arrays contain *exactly* 2 elements. Now i want to find all documents where all of those array elements are elements of my query array.

Example Documents:

`{ a:["1","2"] },`

{ a:["2","3"] },

{ a:["1","3"] },

{ a:["1","4"] }

Query array:

`["1","2","3"]`

The query should find the first 3 documents, but not the last one, since there is no "4" in my query array.

Expected Result:

`{ a:["1","2"] },`

{ a:["2","3"] },

{ a:["1","3"] }

Looking forward to a helpful answer :).

Answer

Since the size is static, you can just check that both elements are in [1,2,3];

```
db.test.find(
{ $and: [ { "a.0": {$in: ["1","2","3"] } },
{ "a.1": {$in: ["1","2","3"] } } ] },
{ _id: 0, a: 1 }
)
>>> { "a" : [ "1", "2" ] }
>>> { "a" : [ "2", "3" ] }
>>> { "a" : [ "1", "3" ] }
```

EDIT: Doing it dynamically is a bit more hairy, I can't think of a way without the aggregation framework. Just count matches as 0 and non matches as 1, and finally remove all groups that have a sum != 0;

```
db.test.aggregate(
{ $unwind: "$a" },
{ $group: { _id: "$_id",
a: { $push: "$a" },
fail: { $sum: {$cond: { if: { $or: [ { $eq:["$a", "1"] },
{ $eq:["$a", "2"] },
{ $eq:["$a", "3"] }]
},
then: 0,
else: 1 } } } } },
{ $match: { fail: 0 } },
{ $project:{ _id: 0, a: 1 } }
)
>>> { "a" : [ "1", "3" ] }
>>> { "a" : [ "2", "3" ] }
>>> { "a" : [ "1", "2" ] }
```