Mongo Basic Drills

Mongo Basic Drills

by John Vincent


Posted on April 15, 2017



Practice CRUD operations on a Mongo shell.

This article refers to a Thinkful course 2.1.3

Mongo Basic Drills

Getting Started

Download and import data into MongoDB.

For details, see this article

Basics

Start with:

use tempTestDb
db (show current database)
show dbs
show tables
show collections (show tables)

db.getCollectionNames()

Get all

Find the command that retrieves all restaurants.

db['restaurants'].find()

db.restaurants.find()

coll = db.getCollection("restaurants")
coll.find()

coll = db.restaurants
coll.find()

Limit and sort

Find the command that the first 10 restaurants that appear when db.restaurants is alphabetically sorted by the name property.

db.restaurants.
  find().
  sort({name: 1}).
  limit(10);

Get by _id

Retrieve a single restaurant by _id from the restaurants collection. This means you’ll first need to get the _id for one of the restaurants imported into the database.

Get the objectid of the first record and then find the record for that objectid

var objectId = db.restaurants.findOne({}, {_id: 1})._id
db.restaurants.find({_id: objectId});

Verify it is the first record:

db.restaurants.findOne()

Another way to find a record by objectid

var documentId = ObjectId('58f25d86e6f2f6c130af4d9f');
db.restaurants.findOne({_id: documentId});

Get by value

Write a command that gets all restaurants from the borough of “Queens”.

db.restaurants.find({borough: "Queens"});

Count these records:

db.restaurants.
  find({borough: "Queens"}).count();
5656

ascending order:

db.restaurants.
  find({borough: "Queens"}).
  sort({name: 1});

descending order:

db.restaurants.
  find({borough: "Queens"}).
  sort({name: -1});

Limit fields:

db.restaurants.
  find({borough: "Queens"}, {borough: 1, name: 1}).
  sort({name: 1});

db.restaurants.
  find({borough: "Queens"}, {borough: 1, name: 1}).
  sort({name: -1});

db.restaurants.
  find({}, {borough: 1, name: 1}).
  sort({name: -1});

db.restaurants.
  find({}, {name: 1, borough: 1}).
  sort({name: -1});

db.restaurants.
  find({}, {name: 1}).
  sort({name: -1});

db.restaurants.
  find({}, {name: 1}).
  sort({name: 1});

Do not list a column:

db.restaurants.
  find({}, {name: 1, _id: 0}).
  sort({name: -1});

Exclude columns:

db.restaurants.
  find({borough: "Queens"}, {borough: 0, name: 0, address: 0, grades: 0}).
  sort({name: -1});

List field in embedded document:

db.restaurants.
  find({borough: "Queens"}, {borough: 1, name: 1, "address.street": 1}).
  sort({name: -1});

db.restaurants.
  find({borough: "Queens"}, {borough: 1, name: 1, "grades.score": 1}).
  sort({name: -1});

Ignore empty string:

db.restaurants.
  find({name: {$ne:""}}, {name: 1}).
  sort({name: 1})

Count

Write a command that gives the number of documents in db.restaurants

db.restaurants.count()

db['restaurants'].count()
=> 25359

db.restaurants.find({name: {$ne:""}}).count()
=> 25208

List the collection and count:

db.getCollectionNames().map(function(name) {
    return { "name": name, "count": db[name].count() }
})

Count by nested value

Write a command that gives the number of restaurants whose zip code value is 11206. Note that this property is at document.address.zipcode, so you’ll need to use dot notation to query on the nested zip code property.

Notice field is in quote "address.zipcode"

db.restaurants.
  find({"address.zipcode": {$eq:"11206"}}).count()
155

With this syntax, the equals operator is assumed:

db.restaurants.
  find({"address.zipcode": "11206"}).count()

List all records with "address.zipcode" = "11206"

db.restaurants.
  find({"address.zipcode": {$eq:"11206"}}, {"address.zipcode": 1})

Delete by id

Write a command that deletes a document from db.restaurants. This means you’ll first need to get the _id for one of the restaurants imported into the database.

Let’s delete the first record with zipcode="11206"

First, get a count of records with zipcode="11206"

db.restaurants.find({"address.zipcode": {$eq:"11206"}}).count()
=> 155

Make objectId

var objectId = db.restaurants.findOne({"address.zipcode": {$eq:"11206"}}, {_id: 1})._id

Ensure it is only one record:

db.restaurants.find({_id: objectId}).count();
=> 1

Delete the record:

db.restaurants.remove({_id: objectId});

Find the record by objectId to ensure it does not exist:

db.restaurants.find({_id: objectId}).count();
=> 0

Count of records with zipcode="11206", should be one less:

db.restaurants.find({"address.zipcode": {$eq:"11206"}}).count()
=> 154

Update a single document

Write a command that sets the name property of a document with a specific _id to 'Bizz Bar Bang'. Make sure that you’re not replacing the existing document, but instead updating only the name property.

Let’s update the first record with zipcode="11206"

var objectId = db.restaurants.findOne({"address.zipcode": {$eq:"11206"}}, {_id: 1})._id
db.restaurants.findOne({_id: objectId});

db.restaurants.updateOne(
  {_id: objectId},
  {$set: {name: "Foo Bar Bizz Bang"}}
);

db.restaurants.findOne({_id: objectId});

Update many documents

Uh oh, two zip codes are being merged! The ‘10035’ zip code is being merged with ‘10036’. Write a command that updates values accordingly.

What am I dealing with? Zip Code by NYC Neighborhoods

Zip codes 10035 and 10036 are both in Manhattan.

Check that all these records really are in borough: manhattan

db.restaurants.find( {
    $and: [
        {"address.zipcode": { $eq: "10035" } },
        {"borough": { $eq: "Manhattan"} }
    ]
}).count()
=> 87

db.restaurants.find( {
    $and: [
        {"address.zipcode": { $eq: "10035" } },
        {"borough": { $ne: "Manhattan"} }
    ]
}).count()
=> 0

db.restaurants.find( {
    $and: [
        {"address.zipcode": { $eq: "10036" } },
        {"borough": { $eq: "Manhattan"} }
    ]
}).count()
=> 610

db.restaurants.find( {
    $and: [
        {"address.zipcode": { $eq: "10036" } },
        {"borough": { $ne: "Manhattan"} }
    ]
}).count()
=> 1

One trouble record, let’s see it:

db.restaurants.find( {
    $and: [
        {"address.zipcode": { $eq: "10036" } },
        {"borough": { $ne: "Manhattan"} }
    ]
})
{ "_id" : ObjectId("58f25d87e6f2f6c130afa02a"), "address" : { "building" : "253", "coord" : [ -73.986403, 40.7602027 ], "street" : "W 47Th St", "zipcode" : "10036" }, "borough" : "Missing", "cuisine" : "Chicken", "grades" : [ { "date" : ISODate("2014-09-30T00:00:00Z"), "grade" : "A", "score" : 7 } ], "name" : "Buffalo Wild Wings Grill & Bar", "restaurant_id" : "50005585" }

Notice that borough = “Missing”

Let’s fix that:

var objectId = db.restaurants.findOne({
    $and: [
        {"address.zipcode": { $eq: "10036" } },
        {"borough": { $ne: "Manhattan"} }
    ]
},
{_id: 1})._id
db.restaurants.findOne({_id: objectId});

object id: 58f25d87e6f2f6c130afa02a

db.restaurants.updateOne(
  {_id: objectId},
  {$set: {borough: "Manhattan"}}
);

db.restaurants.findOne({_id: objectId});

Fixed that. Redo the counts:

db.restaurants.find( {
    $and: [
        {"address.zipcode": { $eq: "10036" } },
        {"borough": { $eq: "Manhattan"} }
    ]
}).count()
=> 611

db.restaurants.find( {
    $and: [
        {"address.zipcode": { $eq: "10036" } },
        {"borough": { $ne: "Manhattan"} }
    ]
}).count()
=> 0

Counts of both zip codes:

db.restaurants.
  find({"address.zipcode": {$eq:"10035"}}).count()
=> 87

db.restaurants.
  find({"address.zipcode": {$eq:"10036"}}).count()
=> 611

As _id is the unique key for documents within the collection, just updating the zip code will act as a merge.

db.restaurants.updateMany(
  {"address.zipcode": {$eq:"10035"}},
  {$set: {"address.zipcode": "10036"}}
);

=> { "acknowledged" : true, "matchedCount" : 87, "modifiedCount" : 87 }

Redo counts of both zip codes:

db.restaurants.
  find({"address.zipcode": {$eq:"10035"}}).count()
=> 0

db.restaurants.
  find({"address.zipcode": {$eq:"10036"}}).count()
=> 698

Count of 698 records is indeed = 87 + 611

End