Fetch Unique Document in Mongodb? - mongodb

Question

Sample Data:
{"_id" : ObjectId("58bd10e4ff1743c527754160"),
"data" : [
{
"No" : "70",
"Type" : "82",
"Device" : "01",
"timestamp" : "2017-03-06 13:00:32"
}]
},
{"_id" : ObjectId("58bd10sdfe4ff1743csdf0754"),
"data" : [
{
"No" : "75",
"Type" : "22",
"Device" : "02",
"timestamp" : "2017-03-06 13:00:32"
}]
}
I have some document which having same timestamp,so I want to find only unique document on the basis of timestamp.
I have done distinct of timestamp but I want full document.
desired Output:
If same timestamp is there I want only One document.

Answer 1

You will only get one output if you run this query.
db.getCollection('tests').aggregate([{$match:{"data.timestamp":"2017-03-06 13:00:32"}},{$limit:1}])

Answer 2

Solution 1:
db.your_collection.aggregate([
{
$group:{
_id:"$data.timestamp",
data:{
$first:"$data"
}
}
}
])
This will give you following :
{ "_id" : [ "2017-03-06 13:00:32" ], "data" : [ { "No" : "70", "Type" : "82", "Device" : "01", "timestamp" : "2017-03-06 13:00:32" }, { "No" : "10", "Type" : "20", "Device" : "01", "timestamp" : "2018-02-04 10:00:00" } ] }
Solution 2 :
db.your_collection.aggregate([
{ $unwind : '$data'},
{ $group : {
_id : '$data.timestamp',
'No': { $first : '$data.No'},
'Type': { $first : '$data.Type'},
'Device': { $first : '$data.Device'},
'timestamp': { $first : '$data.timestamp'},
}
}
]);
This will give you following :
[
{ "_id" : "2017-03-06 13:00:32", "No" : "70", "Type" : "82", "Device" : "01", "timestamp" : "2017-03-06 13:00:32" },
{ "_id" : "2018-02-04 10:00:00", "No" : "10", "Type" : "20", "Device" : "01", "timestamp" : "2018-02-04 10:00:00" },
]