In my MongoDB, I have a student collection with 10 records having fields name and roll. One record of this collection is:
{
"_id" : ObjectId("53d9feff55d6b4dd1171dd9e"),
"name" : "Swati",
"roll" : "80",
}
I want to retrieve the field roll only for all 10 records in the collection as we would do in traditional database by using:
SELECT roll FROM student
I went through many blogs but all are resulting in a query which must have WHERE clause in it, for example:
db.students.find({ "roll": { $gt: 70 })
The query is equivalent to:
SELECT * FROM student WHERE roll > 70
My requirement is to find a single key only without any condition. So, what is the query operation for that.
From the MongoDB docs:
A projection can explicitly include several fields. In the following operation, find() method returns all documents that match the query. In the result set, only the item and qty fields and, by default, the _id field return in the matching documents.
db.inventory.find( { type: 'food' }, { item: 1, qty: 1 } )
In this example from the folks at Mongo, the returned documents will contain only the fields of item, qty, and _id.
Thus, you should be able to issue a statement such as:
db.students.find({}, {roll:1, _id:0})
The above statement will select all documents in the students collection, and the returned document will return only the roll field (and exclude the _id).
If we don't mention _id:0 the fields returned will be roll and _id. The '_id' field is always displayed by default. So we need to explicitly mention _id:0 along with roll.
get all data from table
db.student.find({})
SELECT * FROM student
get all data from table without _id
db.student.find({}, {_id:0})
SELECT name, roll FROM student
get all data from one field with _id
db.student.find({}, {roll:1})
SELECT id, roll FROM student
get all data from one field without _id
db.student.find({}, {roll:1, _id:0})
SELECT roll FROM student
find specified data using where clause
db.student.find({roll: 80})
SELECT * FROM students WHERE roll = '80'
find a data using where clause and greater than condition
db.student.find({ "roll": { $gt: 70 }}) // $gt is greater than
SELECT * FROM student WHERE roll > '70'
find a data using where clause and greater than or equal to condition
db.student.find({ "roll": { $gte: 70 }}) // $gte is greater than or equal
SELECT * FROM student WHERE roll >= '70'
find a data using where clause and less than or equal to condition
db.student.find({ "roll": { $lte: 70 }}) // $lte is less than or equal
SELECT * FROM student WHERE roll <= '70'
find a data using where clause and less than to condition
db.student.find({ "roll": { $lt: 70 }}) // $lt is less than
SELECT * FROM student WHERE roll < '70'
I think mattingly890 has the correct answer , here is another example along with the pattern/commmand
db.collection.find( {}, {your_key:1, _id:0})
> db.mycollection.find().pretty();
{
"_id": ObjectId("54ffca63cea5644e7cda8e1a"),
"host": "google",
"ip": "1.1.192.1"
}
db.mycollection.find({},{ "_id": 0, "host": 1 }).pretty();
Here you go , 3 ways of doing , Shortest to boring :
db.student.find({}, 'roll _id'); // <--- Just multiple fields name space separated
// OR
db.student.find({}).select('roll _id'); // <--- Just multiple fields name space separated
// OR
db.student.find({}, {'roll' : 1 , '_id' : 1 ); // <---- Old lengthy boring way
To remove specific field use - operator :
db.student.find({}).select('roll -_id') // <--- Will remove id from result
While gowtham's answer is complete, it is worth noting that those commands may differ from on API to another (for those not using mongo's shell).
Please refer to documentation link for detailed info.
Nodejs, for instance, have a method called `projection that you would append to your find function in order to project.
Following the same example set, commands like the following can be used with Node:
db.student.find({}).project({roll:1})
SELECT _id, roll FROM student
Or
db.student.find({}).project({roll:1, _id: 0})
SELECT roll FROM student
and so on.
Again for nodejs users, do not forget (what you should already be familiar with if you used this API before) to use toArray in order to append your .then command.
Try the following query:
db.student.find({}, {roll: 1, _id: 0});
And if you are using console you can add pretty() for making it easy to read.
db.student.find({}, {roll: 1, _id: 0}).pretty();
Hope this helps!!
Just for educational purposes you could also do it with any of the following ways:
1.
var query = {"roll": {$gt: 70};
var cursor = db.student.find(query);
cursor.project({"roll":1, "_id":0});
2.
var query = {"roll": {$gt: 70};
var projection = {"roll":1, "_id":0};
var cursor = db.student.find(query,projection);
`
db.<collection>.find({}, {field1: <value>, field2: <value> ...})
In your example, you can do something like:
db.students.find({}, {"roll":true, "_id":false})
Projection
The projection parameter determines which fields are returned in the
matching documents. The projection parameter takes a document of the
following form:
{ field1: <value>, field2: <value> ... }
The <value> can be any of the following:
1 or true to include the field in the return documents.
0 or false to exclude the field.
NOTE
For the _id field, you do not have to explicitly specify _id: 1 to
return the _id field. The find() method always returns the _id field
unless you specify _id: 0 to suppress the field.
READ MORE
For better understanding I have written similar MySQL query.
Selecting specific fields
MongoDB : db.collection_name.find({},{name:true,email:true,phone:true});
MySQL : SELECT name,email,phone FROM table_name;
Selecting specific fields with where clause
MongoDB : db.collection_name.find({email:'you#email.com'},{name:true,email:true,phone:true});
MySQL : SELECT name,email,phone FROM table_name WHERE email = 'you#email.com';
This works for me,
db.student.find({},{"roll":1})
no condition in where clause i.e., inside first curly braces.
inside next curly braces: list of projection field names to be needed in the result and 1 indicates particular field is the part of the query result
getting name of the student
student-details = db.students.find({{ "roll": {$gt: 70} },{"name": 1, "_id": False})
getting name & roll of the student
student-details = db.students.find({{ "roll": {$gt: 70}},{"name": 1,"roll":1,"_id": False})
I just want to add to the answers that if you want to display a field that is nested in another object, you can use the following syntax
db.collection.find( {}, {{'object.key': true}})
Here key is present inside the object named object
{ "_id" : ObjectId("5d2ef0702385"), "object" : { "key" : "value" } }
var collection = db.collection('appuser');
collection.aggregate(
{ $project : { firstName : 1, lastName : 1 } },function(err, res){
res.toArray(function(err, realRes){
console.log("response roo==>",realRes);
});
});
it's working
Use the Query like this in the shell:
1. Use database_name
e.g: use database_name
2. Which returns only assets particular field information when matched , _id:0 specifies not to display ID in the result
db.collection_name.find( { "Search_Field": "value" },
{ "Field_to_display": 1,_id:0 } )
If u want to retrieve the field "roll" only for all 10 records in the collections.
Then try this.
In MongoDb :
db.students.find( { } , { " roll " : { " $roll " })
In Sql :
select roll from students
The query for MongoDB here fees is collection and description is a field.
db.getCollection('fees').find({},{description:1,_id:0})
Apart from what people have already mentioned I am just introducing indexes to the mix.
So imagine a large collection, with let's say over 1 million documents and you have to run a query like this.
The WiredTiger Internal cache will have to keep all that data in the cache if you have to run this query on it, if not that data will be fed into the WT Internal Cache either from FS Cache or Disk before the retrieval from DB is done (in batches if being called for from a driver connected to database & given that 1 million documents are not returned in 1 go, cursor comes into play)
Covered query can be an alternative. Copying the text from docs directly.
When an index covers a query, MongoDB can both match the query conditions and return the results using only the index keys; i.e. MongoDB does not need to examine documents from the collection to return the results.
When an index covers a query, the explain result has an IXSCAN stage that is not a descendant of a FETCH stage, and in the executionStats, the totalDocsExamined is 0.
Query : db.getCollection('qaa').find({roll_no : {$gte : 0}},{_id : 0, roll_no : 1})
Index : db.getCollection('qaa').createIndex({roll_no : 1})
If the index here is in WT Internal Cache then it would be a straight forward process to get the values. An index has impact on the write performance of the system thus this would make more sense if the reads are a plenty compared to the writes.
If you are using the MongoDB driver in NodeJs then the above-mentioned answers might not work for you. You will have to do something like this to get only selected properties as a response.
import { MongoClient } from "mongodb";
// Replace the uri string with your MongoDB deployment's connection string.
const uri = "<connection string uri>";
const client = new MongoClient(uri);
async function run() {
try {
await client.connect();
const database = client.db("sample_mflix");
const movies = database.collection("movies");
// Query for a movie that has the title 'The Room'
const query = { title: "The Room" };
const options = {
// sort matched documents in descending order by rating
sort: { "imdb.rating": -1 },
// Include only the `title` and `imdb` fields in the returned document
projection: { _id: 0, title: 1, imdb: 1 },
};
const movie = await movies.findOne(query, options);
/** since this method returns the matched document, not a cursor,
* print it directly
*/
console.log(movie);
} finally {
await client.close();
}
}
run().catch(console.dir);
This code is copied from the actual MongoDB doc you can check here.
https://docs.mongodb.com/drivers/node/current/usage-examples/findOne/
db.student.find({}, {"roll":1, "_id":0})
This is equivalent to -
Select roll from student
db.student.find({}, {"roll":1, "name":1, "_id":0})
This is equivalent to -
Select roll, name from student
In mongodb 3.4 we can use below logic, i am not sure about previous versions
select roll from student ==> db.student.find(!{}, {roll:1})
the above logic helps to define some columns (if they are less)
Using Studio 3T for MongoDB, if I use .find({}, { _id: 0, roll: true }) it still return an array of objects with an empty _id property.
Using JavaScript map helped me to only retrieve the desired roll property as an array of string:
var rolls = db.student
.find({ roll: { $gt: 70 } }) // query where role > 70
.map(x => x.roll); // return an array of role
Not sure this answers the question but I believe it's worth mentioning here.
There is one more way for selecting single field (and not multiple) using db.collection_name.distinct();
e.g.,db.student.distinct('roll',{});
Or, 2nd way: Using db.collection_name.find().forEach(); (multiple fields can be selected here by concatenation)
e.g., db.collection_name.find().forEach(function(c1){print(c1.roll);});
_id = "123321"; _user = await likes.find({liker_id: _id},{liked_id:"$liked_id"}); ;
let suppose you have liker_id and liked_id field in the document so by putting "$liked_id" it will return _id and liked_id only.
For Single Update :
db.collection_name.update({ field_name_1: ("value")}, { $set: { field_name_2 : "new_value" }});
For MultiUpdate :
db.collection_name.updateMany({ field_name_1: ("value")}, { $set: {field_name_2 : "new_value" }});
Make sure indexes are proper.
Related
Trying so that I can check if a user has previously liked a recipe so they can't like it twice.
I have users in my DB that look like:
username:"username"
password:"password"
likes:Array
0:"recipe_id"
Im trying to use the query:
if mongo.db.users.find_one({'username': session.get('USERNAME')}, {'likes': [recipe_id]}) == None:
mongo.db.recipe.update_one({'_id': ObjectId(recipe_id)}, {
'$inc': {'likes': 1}})
mongo.db.users.update_one({'username': session['USERNAME']}, {
'$push': {'likes': recipe_id}})
but its just returning regardless of the recipeId being in the array or not. I think this is because its doing two queries. Is there a way to specify the second query to only look at that specific user ?
Issue with your query is you're having likes : [recipe_id] in projection part of .find_one. So it has to be in filter part, Basically second argument to .find() or .find_one() is projection which helps to project certain fields in the document.
Try this query :
/** So all you need is to check if `recipe_id` exists in `likes` array or not,
* then if it exists instead of returning entire doc,
* using projection just return `_id` of doc which is quiet enough for your need,
* in other case query will return none if no matching doc exists */
if mongo.db.users.find_one({'username': session.get('USERNAME'), 'likes': recipe_id}, {_id :1}) == None:
mongo.db.recipe.update_one({'_id': ObjectId(recipe_id)}, {
'$inc': {'likes': 1}})
mongo.db.users.update_one({'username': session['USERNAME']}, {
'$push': {'likes': recipe_id}})
In .find_one() you don't need to wrap recipe_id in array like what you're doing : likes : [recipe_id] is not needed, If you wanted to do it you need to use $in operator like this { likes : {$in : [recipe_id] } } in filter part of query.
Ref : .findOne() & .find_one()
Excuse my ignorance, I am new in MongoDB. I am having tree collections, where the one is a superset of the other two whose elements are not overlapped. Each item is distinguish by a unique string id. What I want is to get the items of the superset that are not included in the other two collections. Could you please provide me some hint on how do do this efficiently?
Thanks.
EDIT:
Superset structure:
{ "_id" : 1, "str_id" : "ABC1fd3fsewer", "date": "a day" }
Subset 1 structure: { "_id" : 1, "str_id" : "ABre1fd3fsewer", "description" : "product" }
Subset 2 structure: { "_id" : 1, "str_id" : "ABC1fd3fsewfe"}
Each collection has a different structure but all have a common filed, the str_id.
EDIT Improved by #Neel suggestion
I have following format:
parent = [{'str_id':'a', 'tag1':'parent_random', 'tag2': 'parent_random', 'tag3':'parent_random'},{'str_id':'b',...},{'str_id':'c',...},{'str_id':'d',...}...]
child1 = [{'str_id':'a', 'tag2': child1_random'},{'str_id':'b', 'tag2': 'child1_random'}]
child2 = [{'str_id':'c', 'tag1':'child2_random'}]
and I want
outcome = [{'str_id':'c', 'tag1':'parent_random', 'tag2': 'parent_random', 'tag3':'parent_random'},{'str_id':'d', 'tag1':'parent_random', 'tag2': 'parent_random', 'tag3':'parent_random'}]
It sounds like you'll need an aggregate operation.
This document might help you:
Lookup in an array
You can do multiple lookups with one aggregate operation so you can check both the subset collections.
I am going to assume you are working with a REST API and that the client is sending a request for a subset of documents from the superset collection. You can send the array of documents you want to check from superset from the client then:
1 - match all the documents in superset to the array of documents you're sending
2 - unwind your superset document array
3 - lookup the subset collections on "str_id" field and set to a field, like "subset_one_results".
4 - do a match operation on both subset results that returns an empty array on, say, "subset_one_results"... this will match all superset documents that are not contained in subset1 for example.
$match({ $and : { "subset_one_results" : { $eq : [] } }, { "subset_two_results" : { $eq : [] } } })
5 - group them in a new array if you want to return them as an array to the client.
To increase the performance of your operations, you have to determine how often this request will be made. If it risks being often, be sure to create an index on the field that will be solicited if it's not an ObjectId field. I can't tell from your code if you are using a custom string field or an ObjectId, which is why I'm bringing up this point.
I don't know what you're using for making your queries (pure MongoDB query language, driver, etc.) so I am not sure how to answer with code hence delineating the steps up above.
I'm new to programming and I've been learning Flask for about a week now, trying to design an e-commerce website.
I have a collection called users, in which each entry looks like this
{
"_id" : ObjectId("5b9170f1c5eb754e3ab8917f"),
"username" : "suhas",
"password" : "suhas",
"email" : "suhas",
"account_type" : "buyer",
"cart" : [
{
"product_id" : "5b915dd3c5eb754278e160e7",
"quantity" : 7
},
{
"product_id" : "5b915e3fc5eb754278e160e8",
"quantity" : 3
}
]
}
and a users.products collection, one of the elements look like this
{
"_id" : ObjectId("5b915f02c5eb7544108a14b9"),
"product name" : "Laptop",
"price" : 50000,
"description" : "HP laptop",
"user_id" : "5b914fc3c5eb753eaf81770c",
"username" : "chiranth"
}
I wanted to add a feaure "add to cart" that adds the product_id and quantity into a cart(type list) as dictionaries, as seen in the db.users collection.
The code in Python I've written as:
cart_dict = user_info.get("cart")
for dict1 in cart_dict:
if dict1["product_id"]==product_id:
db["users"].update({"_id" : ObjectId(user_id),"cart.product_id":product_id},{ '$inc':{ 'cart.$.quantity':quantity}})
break
else:
db["users"].update({"_id":ObjectId(user_id)},{"$addToSet":{"cart":{"$each":[{"product_id":product_id,"quantity":quantity}]}}})
The problem is, however, that if I increment the second product by adding 2 more to the cart adding to the present 3, the if condition checks the first product, sees that product ID doesn't match and just adds it as a new product AND THEN moves to product 2.
So my question would be: how do I ask the if condition to check all product_id's first AND THEN, if none match, add the product?
EDIT : to clarify,
my issue is with the IF condition. I want it to first compare product A, if that isnt satisfied, move to product B, and IF neither match, THEN it should go to the else loop. How do i do that?
Hey the code you have provided, their is unnecessary second for loop on same data just above the if condition.
you should remove that line. As pr updating data the code i used and works is as :
collection.update({'_id':int(n)},{'$set' : {'CloseDate':TodayDate,'Status':'close'}})
in this the where is provided in _id and the fields CloseDate & Status are being updated, this works, so you should put your new values that are to be updated in DB in some variables and directly assign them in your documents name.
Hope this helps.
I have a database for a backup service I'm writing to backup Yahoo! Groups. It incrementally retrieves messages, which have a contiguous numeric id. stored in a 'message_id' field. So, if the last message on the service is message number 10000, then once the backup is complete, the database should contain 10000 documents, with the sorted 'message_id's of each document being equivalent to range(1, 10000+1).
I'd like to write a query yielding the missing message ids. So if I have 9995 documents in the database, and messages 10, 15, 49, 99, and 1043 are missing, it should return [10, 15, 49, 99, 1043].
I've done the following, getting just the ids from the database and running a set intersection in my app code:
def missing_message_ids(self):
"""Return the set of the ids of all missing messages.."""
latest = self.get_latest_message()
ids = set(range(1, latest['_id']+1))
present_ids = set(doc['_id'] for doc in self.db.messages.find({}, {'_id': 1}))
return ids - present_ids
This is fine for my purposes, but it seems like it might get too slow for a vast number of messages. This is more for curiosity's sake than a real performance requirement: Is there any more efficient way to do this, perhaps entirely on the database engine?
in SQL word one could use CTE for that, in mongo we can use aggregation with $lookup as a kind of CTE (common table expressions)
having this data structure
{
"_id" : ObjectId("575deea531dcfb59af388e17"),
"mesId" : 4.0
}, {
"_id" : ObjectId("575deea531dcfb59af388e18"),
"mesId" : 6.0
}
with missing "mesId" : 5.0 we can use this aggregation query, which will project all next expected ids, and join on them. The limitation here is if we have missing more than one message in sequence, but this could be extended by projecting next Id and making $lookup again.
var project = {
$project : {
_id : 0,
mesId : 1,
nextId : {
$sum : ["$mesId", 1]
}
}
}
var lookup = {
$lookup : {
from : "claudiu",
localField : "nextId",
foreignField : "mesId",
as : "missing"
}
}
var match = {
$match : {
missing : []
}
}
db.claudiu.aggregate([project, lookup, match])
and output:
{
"mesId" : 4.0,
"nextId" : 5.0,
"missing" : []
}
I have a mongo collection with multiple documents, suppose the following (assume Tom had two teachers for History in 2012 for whatever reason)
{
"name" : "Tom"
"year" : 2012
"class" : "History"
"Teacher" : "Forester"
}
{
"name" : "Tom"
"year" : 2011
"class" : "Math"
"Teacher" : "Sumpra"
}
{
"name" : "Tom",
"year" : 2012,
"class" : "History",
"Teacher" : "Reiser"
}
I want to be able to query for all the distinct classes "Tom" has ever had, even though Tom has had multiple "History" classes with multiple teachers, I just want the query to get the minimal number of documents such that Tom is in all of them, and "History" shows up one time, as opposed to having a query result that contains multiple documents with "History" repeated.
I took a look at:
http://mongoengine-odm.readthedocs.org/en/latest/guide/querying.html
and want to be able to try something like:
student_users = Students.objects(name = "Tom", class = "some way to say distinct?")
Though it does not appear to be documented. If this is not the syntactically correct way to do it, is this possible in mongoengine, or is there some way to accomplish with some other library like pymongo? Or do i have to query for all documents with Tom then do some post-processing to get to unique values? Syntax would be appreciated for any case.
First of all, it's only possible to get distinct values on some field (only one field) as explained in MongoDB documentation on Distinct.
Mongoengine's QuerySet class does support distinct() method to do the job.
So you might try something like this to get results:
Students.objects(name="Tom").distinct(field="class")
This query results in one BSON-document containing list of classes Tom attends.
Attention Note that returned value is a single document, so if it exceeds max document size (16 MB), you'll get error and in that case you have to switch to map/reduce approach to solve such kind of problems.
import pymongo
posts = pymongo.MongoClient('localhost', 27017)['db']['colection']
res = posts.find({ "geography": { "$regex": '/europe/', "$options": 'i'}}).distinct('geography')
print type(res)
res.sort()
for line in res:
print line
refer to http://docs.mongodb.org/manual/reference/method/db.collection.distinct/
distinct returns a list , will be printed on print type(res) , you can sort a list with res.sort() , after that it will print the values of the sorted list.
Also you can query posts before select distinct values .
student_users = Students.objects(name = "Tom").distinct('class')