Find values in mongodb with array

Question

I am trying to get values from a mongodb table which stores data about stock taking dates in a shop(the date is stored as string) along with python.

the data stored in the table:

{
    "_id" : ObjectId("5645f0443f32df13e0dc786e"),
    "RequiredDate" : "28-9-2015",
    "Name": "xyz"
}
{
    "_id" : ObjectId("5645f0c73f32df062064a0f6"),
    "RequiredDate" : "29-9-2015",
    "Name": "abc"
}

I would like to find the name of the products which are required on both days 28-9-2015 & 29-9-2015.

Right now do it individually like this

db.stockdate.find({"RequiredDate": "28-9-2015"})
db.stockdate.find({"RequiredDate": "29-9-2015"})

Can i do these in a single query, Is there any option for that? like giving both date in a single query and get the same result which i got with two query.

ZeMoon · Accepted Answer · 2015-11-14 13:24:09Z

1

Use the $in operator.

db.stockdate.find({"RequiredDate": {$in: ["28-9-2015", "29-9-2015"]}})

Additionally, if you'd just like the names of the products, you can use

db.stockdate.distinct("Name", {"RequiredDate": {$in: ["28-9-2015", "29-9-2015"]}})

If you want to group the records based on the dates, you will need to use aggregation:

db.stockdate.aggregate([
    {$match: {"RequiredDate": {$in: ["28-9-2015", "29-9-2015"]}}},
    {$group: {_id: "RequiredDate", Names: {$push: "$Name"}, count: {$sum: 1}}}
])

The above aggregation query will produce a result like this:

{
    "_id" : "28-9-2015",
    "Names": ["xyz"],
    "count": 1
}
{
    "_id" : "29-9-2015",
    "Names": ["abc", "def"],
    "count": 2
}

You can add as many dates to the array, and this will return records that have the reuiredDate set to any of those dates.

edited Nov 14, 2015 at 13:24

answered Nov 14, 2015 at 13:09

ZeMoon

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2 Comments

user5189062 Over a year ago

There might me multiple results for the above query with(multiple dates) is there a way to know which date has stock taking so that i can give a total count of all the products that they are stock taking?

ZeMoon Over a year ago

Okay. For that you will need to use aggregation. Gimme a minute for that.

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