I am a new user of Mongo Db, please let me know the difference in executing both queries - mongodb

Hi I found that the output of the two queries was same, but I want to know is there any difference in executing the query.
First:
db.collectionname.find({}).pretty()
Second:
db.offers.find().pretty()

There is no difference between these two queries.
db.collectionname.find({}).pretty()
You are not giving any query params here so the results are same like
db.collectionname.find().pretty()

In short: Both function in the same way.
1) db.collectionname.find({})
Here you are not specifying any query parameters and its just an empty document {} so it will return you all the documents present in the collection
2) db.offers.find()
Here you didn't specify any queries. So it need not even look at the parameters, it will just print you all the documents in the collection. find() short form of find({})

Related

Pymongo - find multiple different documents

my question is very similar to how-to-get-multiple-document-using-array-of-mongodb-id, however, I would like to find multiple documents without using the _id.
That is, consider that I have documents such as
the
document = { _id: _id, key_1: val_1, key_2: val_2, key_3: val_3}
I need to be able to .find() by multiple parameters, as for example,
query_1 = {key_1: foo, key_2: bar}
query_2 = {key_1: foofoo, key_2: barbar}
Right now, I am running a query for query_1, followed by a query for query_2.
As it turns out, this method is extremely inefficient.
I tried to add concurrency as to make it faster, but the speedup was not even 2x.
Is it possible to query multiple documents at once?,
I am looking for a method that returns the union of the matches for query_1 AND query_2.
If this is not possible, do you have any suggestions that might speed a query of this type?
Thank you for your help.

find_one query returns just the fields instead of an entry

I'm currently trying to use pymongo's find_one query. When I run the Mongo Shell and execute a findOne query, it get a document that is returned. However when I try using pymongo's find_one query, I always seem to get just the field names instead of an actual entry.
#app.route("/borough/manhattan/")
def manhattan():
restaurantmanhattan = restaurants.find_one({'borough':'Manhattan'})
json_restaurantmanhattan = []
for restaurant in restaurantmanhattan:
json_restaurantmanhattan.append(restaurant)
json_restaurantmanhattan = json.dumps(json_restaurantmanhattan)
return json_restaurantmanhattan
Once I navigate to http://0.0.0.0:5000/borough/manhattan/ I get the following:
["cuisine","borough","name","restaurant_id","grades","address","_id"]
I believe I should be seeing a document entry that meets the query that it has Manhattan listed in the borough.
I'm at a loss as to how I should be writing the query to return that.
Can anyone explain what I'm seeing?
There are many things wrong with your view.
First as you may already know, find_one return a single document as Python dictionary. So in your for loop, you iterating the dictionary keys.
You really do not need that for loop.
import json
#app.route("/borough/manhattan/")
def manhattan():
restaurant_manhattan = restaurants.find_one({'borough':'Manhattan'})
return json.dumps(restaurant_manhattan)

Mongo DB search based on multiple conditions

I am trying to search based on multiple conditions which works but the problem is that does not behave like this.
Assuming i have a search query like
Orders.find({$or: {"status":{"$in":["open", "closed"]},"paymentStatus":{"$in":["unpaid"]}}}
)
and i add another filter parameter like approvalStatus it does not leave the previously found items but rather it treats the query like an AND that will return an empty collection of items if one of the queries does not match.
How can i write a query that regardless of what is passed into it, it will retain previously found items even if there is no record in one of the conditions.
like a simple OR query in sql
I hope i explained this well enough
Using $or here is the right approach, but its value needs to be an array of query expressions, not an object.
So your query should look something like this instead:
Orders.find({$or: [
{"status": {"$in": ["open", "closed"]}},
{"paymentStatus": {"$in": ["unpaid"]}},
{"approvalStatus": {"$in": ["approved"]}}
]})

Mongo Get Count While Returning Whole Documents and Should Queries

I am new to Mongo and can't seem to figure out the following after reading posts and the documentation. I am executing the following query:
db.collection.find({'name':'example name'})
Which returns 14 results. I can get the count of correctly by executing:
db.collection.find({'name':'example name'}).count()
However, I want to return the full documents and the count in a single query, similar to the way Elasticsearch does. Is there anyway to do this.
Additionally, is there any equivalence to Elasticsearch's Bool should query (http://www.elasticsearch.org/guide/en/elasticsearch/reference/current/query-dsl-bool-query.html). Essentially I would want to rank the results, so that those with attribute 'onSale=True' are returned before 'onSale=False'.
I'm not sure about your second question, whether MongoDB provides some mechanism equivalent to Elasticsearch's Bool should query.
But for your 1st question, I think you can use Cursor.
var cursor = db.collection.find({'name':'example name'});
Once you've got the cursor, you can use it for getting the count in the following way:
cursor.count()
as well as for getting the documents wrapped in an array in the following way:
cursor.toArray()
For more info on cursor, please see the below mentioned link:
http://docs.mongodb.org/manual/tutorial/iterate-a-cursor/

MongoDB: How to execute a query to result of another query (nested queries)?

I need to apply a set of filters (queries) to a collection. By default, the MongoDB applies AND operator to all queries submitted to find function. Instead of whole AND I need to apply each query sequentially (one by one). That is, I need to run the first-query and get a set of documents, run the second-query to result of first-query, and so on.
Is this Possible?
db.list.find({..q1..}).find({..q2..}).find({..q3..});
Instead Of:
db.list.find({..q1..}, {..q2..}, {..q3..});
Why do I need this?
Bcoz, the second-query needs to apply an aggregate function to result of first-query, instead of applying the aggregate to whole collection.
Yes this is possible in MongoDB. You can write nested queries as per the requirement.Even in my application I created nested MongoDb queries.If you are familiar with SQL syntax then compare this with in of sql syntax:
select cname from table where cid in (select .....)
In the same way you can create nested MongoDB queries on different collections also.