mongodb javascript：返回仅包含匹配子文档的文档

Question

anyone know how to return the document with only matched sub-document in javascript?

eg here is the database records:

[
  {"name":"bestbuy",notes:["article IT", "article 2"]},
  {"name":"microsoft",notes:["article IT", "another IT", "article 5"]},
  {"name":"IBM",notes:["article 8", "article 9"]}
]

here is my query:

collection.find({"company.notes":/IT/}, function(err,result){})

result is:

[
  {"name":"bestbuy",notes:["article IT", "article 2"]},
  {"name":"microsoft",notes:["article IT", "another IT", "article 5"]},
]

but my expected result is:

[
  {"name":"bestbuy",notes:["article IT"]},
  {"name":"microsoft",notes:["article IT", "another IT"]}
]

any idea? thanks

Answer 1

You can use aggregation

db.collection.aggregate([
    {$match: {"notes": /IT/}}, 
    {$unwind: "$notes"}, 
    {$match: {notes: /IT/}}, 
    {$group: {_id: '$_id', name: {$first: '$name'}, notes: {$push: '$notes'}}},
    {$project: {'name': 1, 'notes': 1, _id: 0}}
])

yields:

{ "name" : "microsoft", "notes" : [ "article IT", "another IT" ] }
{ "name" : "bestbuy", "notes" : [ "article IT" ] }