like:
{id:4563214321,updateTime:long("124354354")}
there are always new collections enter the db, so I would like to always get latest updated documents aka the largest update time. how to design the shell script? thanks in advance.
You can use a combination of limit and sort to achieve this goal.
db.collectionName.find({}).sort({"updateTime" : -1}).limit(1)
This will sort all of your fields based on update time and then only return the one largest value.
I would recommend adding an index to this field to improve performance.
This is a repeated question, you can find an answer in this link
Using findOne in mongodb to get element with max id
use like this,
db.collection.find().sort({updateTime:-1}).limit(1).pretty()
as findOne you can do it with this syntax:
db.collection.findOne({$query:{},$orderby:{_updateTime:-1}})
Related
I am trying to sort the result of my mongoDB query and add a ranking based on that sorting. Currently I only call .find().sort({total: 1}) and this gives me the correct ordering of the documents. But is it possible to "add a field" based on that sorting (basically a ranking field, starting from 1 and counting up)? I tried googling but didnt found anything that suits for this purpose.
Thanks in advance.
Is there a simple OR elegant method (or query that I can write) to retrieve the last updated timestamp (of the last updated document) in a collection. I can write a query like this to find the last inserted document
db.collection.find().limit(1).sort({$natural:-1})
but I need information about the last updated document (it could be an insert or an update).
I know that one way is to query the oplog collection for the last record from a collection. But it seems like an expensive operation given the fact that oplog could be of very large size (also not trustworthy as it is a capped collection). Is there a better way to do this?
Thanks!
You could get the last insert time same way you mentioned in the question:
db.collection.find().sort({'_id': -1}).limit(1)
But, There isn't any good way to see the last update/delete time. But, If you are using replica sets you could get that from the oplog.
Or, you could add new field in document as 'lastModified'.
You can also checkout collection-hooks. I hope this will help
One way to go about it is to have a field that holds the time of last update. You can name it updatedAt. Every time you make an update to the document, you'll just update the value to the current time. If you use the ISO format to store the time, you'll be able to sort without issues (that's what I use).
The other way is the _id field.
Method 1
db.collection.find().limit(1).sort({updatedAt: -1})
Method 2
db.collection.find().limit(1).sort({_id: -1})
You can try with ,
db.collection.findOne().sort({$natural:-1}).limit(1);
I'm writing a long-polling script to check for new documents in my mongo collection and the only way I know of checking to see whether or not changes have been made in the past iteration of my loop is to do a query getting the last document's ID and parsing out the timestamp in that ID and seeing if it's greater than the timestamp left since I last made a request.
Is there some kind of approach that doesn't involve making a query every time or something that makes that query the fastest?
I was thinking a query like this:
db.chat.find().sort({_id:-1}).limit(1);
But it would be using the PHP driver.
The fastest way will be creating indexes on timestamp field.
Creating index:
db.posts.ensureIndex( { timestamp : 1 } )
Optimizes this query:
db.posts.find().sort( { timestamp : -1 } )
findOne give you only one the last timestamp.
nice to help you.
#Zagorulkin Your suggestion is surely going to help in the required scenario. However i don't think so sort() works with findOne().
Is there any way to find last update Document in Collection? in other way sort collection by update
somethings like this
people = Person.objects.order_by_update()
or i must add update time for each doc?
I use mongodb, mongoengine, flask
You must add a field such as last_updated_time if you want to be able to sort in this way. Also, since you're sorting on it, you should probably index it.
The only thing that mongodb stores by default is _id, which can be used roughly as a created_time timestamp.
For example given the BlogPost/Comments schema here:
http://mongoosejs.com/
How would I find all posts with more than five comments? I have tried something along the lines of
where('comments').size.gte(5)
But I'm getting tripped up with the syntax
MongoDb doesn't support range queries with size operator (Link). They recommend you to create a separate field to contain the size of the list that you increment yourself.
You cannot use $size to find a range of sizes (for example: arrays with more than 1 element). If you need to query for a range, create an extra size field that you increment when you add elements.
Note that for some queries, it may be feasible to just list all the counts you want in or excluded using (n)or conditions.
In your example, the following query will give all documents with more than 5 comments (using standard mongodb syntax, not mongoose):
db.col.find({"comments":{"$exists"=>true}, "$nor":[{"comments":{"$size"=>4}}, {"comments":{"$size"=>3}}, {"comments":{"$size"=>2}}, {"comments":{"$size"=>1}}, {"comments":{"$size"=>0}}]})
Obviously, this is very repetitive, so it only makes sense for small boundaries, if at all. Keeping a separate count variable, as recommended in the mongodb docs, is usually the better solution.
It's slow, but you could also use the $where clause:
db.Blog.find({$where:"this.comments.length > 5"}).exec(...);