查询数组中的重复元素（MongoDB）

Question

I have a mongodb collection with an array field, containing a list of strings. There may be repeats in those strings. for example:

doc1 = {a: ["p", "q", "r", "p", "r"]}
doc2 = {a: ["p", "q", "q"]}
doc3 = {a: ["p"]}
doc4 = {a: ["p", "r", "r"]}

I want to find all the documents that, given a string (say, "p"), finds all the documents that have the string at least two times in the array.

For example:

query("p") == [doc1]
query("q") == [doc2]
query("r") == [doc1, doc4]

Is there a way to do this directly in mongo? I know I can query for occurrence once, and then filter the results on my application, but I'd rather avoid that.

Answer 1

You could try something like below. This query returns the _id of the documents matching your query and also the count.

db.mycoll.aggregate([
    {$unwind:"$a"}, 
    {$group:{_id:{_id:"$_id", a:"$a"}, count:{$sum:1}}}, 
    {$match:{"_id.a":"r", count:{$gte:2}}}, 
    {$project:{_id:0, id:"$_id._id", count:1}}
])

Note that $match phase contains "p". You can substitute that with "q" or "r"

Answer 2

var search = 'r';
docs.aggregate([
  {$match: { a : search } }, //step 1, filter to the arrays we care about for speed
  //could do a project here to trim fields depending on object size
  {$unwind: '$a'}, //unwind to create a separate row for each letter
  { $group: { _id: '$_id', total: { $sum: { $cond : [ { $eq: ['$a', search] }, 1, 0] } } } }, //the real work, explained below
  {$match : {total : {$gte: 2} } } //grab the summed items with at least 2
  {$project: {_id: 1} } //grab just the _id field
]  )

Notes:

I believe $elemMatch won't work as it always finds the first item in the array, not every item in the array.

The real work happens in the $group call, where the $sum is based on the condition of finding the element you're searching for in the array. This works because we've unwound them to be separate rows.

Enjoy!