AGGREGATE Movies Tutorial
From NoSQLZoo
Investigating the Movie Database
The movie database includes thousands of movies with documents such as:
{
"_id" : 10522,
"title" : "Alien",
"yr" : 1979,
"director" : "Ridley Scott",
"budget" : 11000000,
"gross" : 104931801,
"cast" : [
"Tom Skerritt",
"Sigourney Weaver",
"Veronica Cartwright",
"Harry Dean Stanton",
"John Hurt",
"Ian Holm",
"Yaphet Kotto"
]
}
Casablanca
Show the details of the movie Casablanca
db.movies.aggregate([
{$match:{
title:'Alien'
}},
]).pretty()
db.movies.aggregate([
{$match:{
title:'Casablanca'
}}
]).pretty()
Tom Hanks
You can use the match operator on listed items like cast. As there is an index on cast these queries operate quickly and do not have to scan the entire collection.
Show the title and year of the 10 most recent Tom Hanks movies - show the most recent first.
db.movies.aggregate([
{$match:{
cast:'Harrison Ford'
}},
{$project:{_id:0,title:1}}
]).pretty()
db.movies.aggregate([
{$match:{
cast:'Tom Hanks'
}},
{$project:{title:1,yr:1,_id:0}},
{$sort:{yr:-1}},
{$limit:10}
])
Dark Years
A "Dark Year" is one in which Jack Black made more than two movies.
List the years in which Jack Black made three or more movies.
One way to do this is to set up an aggregate pipeline with the following stages:
- Use $match to select the Jack Black movies
- Use $group together with $sum:1 to get the years together with a count, you will have to name the count field
- Use $match to get those where the count is $gte:3
- Use $project to extract the year only
- Use sort to get the output in ascending order
db.movies.aggregate([
{$match:{
cast:'Harrison Ford'
}},
{$project:{_id:0,title:1}}
]).pretty()
db.movies.aggregate([
{$match:{
cast:'Jack Black'
}},
{$group:{_id:'$yr',num:{$sum:1}}},
{$match:{num:{$gte:3}}},
{$project:{_id:1}},
{$sort:{_id:1}}
]).pretty()