I am new in MongoDB. I have a collection in MongoDB. I would like to shift a column in collection of MongoDB. I would like to place image as last column and email as before the last column. How can I do that ?
I agree with others who mentioned that the order of keys in a document should not matter.
But if you still want to learn more about it, it is mentioned in the docs that:
MongoDB preserves the order of the document fields following write
operations...
Considering that, you can actually alter the order of fields by removing and re-inserting them (using $unset and $set) or by using the $rename operator which does exactly that. But you will need to do a couple of operations: First rename the image field to something else, than re-rename it back to image like so:
db.test.updateMany({}, {$rename: {image: 'image_'}})
db.test.updateMany({}, {$rename: {image_: 'image'}})
Since this will actually re-insert the image field, it will cause that field to be last in the document.
Related
The chat app schema that I have is something like below.
1. conversations {participants[user_1, user_2], convsersation_id}
2. messages {sender: user_1, sonversation_id, timestamps}
I want to map this relationship using existing _id:ObjectId which is already indexed.
But if I want to get all conversation of user_1 I have to first search in which conversation that user is involed and get that conversation's _id and again search for the messages in messages using that conversation _id.
So my questions are -
Does length of indexed field (here _id) matters while searching?
Should I create another shorter indexed fields?.
Also if there is any better alternative schema please suggest.
I would suggest you to maintain the data as sub documents instead of array. The advantage you have is you can build another index (only) on conversation_id field, which you want to query to know the user's involvement
When you maintain it as array, you cannot index the converstaion_id field separately, instead you will have to build a multi key index, which indexes all the elements of the array (sender and timestamps fields) which you are never going to use for querying and it also increases the index size
Answering you questions:
Does length of indexed field (here _id) matters while searching? - Not really
Should I create another shorter indexed fields? - Create sub-document and index converstaion_id
Also if there is any better alternative schema please suggest. - Maintain the array fields as sub-documents
I need to find all the documents in mongodb that have keywords that start with number 1-9, then add a '+' in front of the keyword, I can easily find the documents but cannot figure out how to update them.
I tried this one, but it doesn't work
db.placements.update({program_id:{$in:[113,107]},
keyword:{$regex:'^[0-9]', $options:'i'}},
{keyword:"+"+$keyword})
It cannot recognize $keyword, I also tried '.keyword', 'keyword', none of them works. Is there any way to reference the document itself like Java does, using 'this', so I can do something like
this.keyword: "+" + this.keyword
You'll have to use the $set operator in the update query to update a specific field. Also, you cannot concatenate string within an update query. One way to do this would be using cursor forEach() in the shell:
db.placements.find({program_id:{$in:[113,107]}, keyword:{$regex:'^[0-9]', $options:'i'}})
.forEach(function(doc){
db.placements.updateOne({_id:doc._id}, {$set:{"keyword":"+" + doc.keyword}})
})
No, you cannot reference a value on the document itself when querying like you can with SQL.
I would suggest querying the document, updating it on your web/app server and then updating the value back to mongodb.
You will also find that your update command above will wipe your entire document leaving only the keyword field. You should use the $set modifier to update a field or set of fields.
db.placements.update(
{
program_id:{$in:[113,107]},
keyword:{$regex:'^[0-9]', $options:'i'}
},
{ $set: {keyword: new_value}})
I have a dictionary. I need to insert column 2 into mongodb corresponding to column 1(key).
Say this is the dictionary:
values = {'a':['1','2','3'],
'b':['1','2'],
'c':['3','4'] }
Right now I am doing this:
for k,v in values.items():
col4.update({"name":k},{"$set":{"fieldName":v}})
But this takes 3 accesses to the db. Is it possible to do it one go like the way $in works.
In your code you are finding each document by name field and set its fieldName to v. There is no update operation in Mongo that can do such thing in one shot for multiple documents.
However there is a bulk insert statement which can be more efficient than multiple inserts or updates. http://docs.mongodb.org/manual/core/bulk-inserts/.
I thinks I previously didn't quite understand what you were asking and wrote the answer below, but I'm still not sure what you mean by $in. Perhaps you can provide example of data before and after update in DB, that way it will be absolutely clear what you are trying to achieve.
OLD answer ... (I'll edit it soon)
You need to restructure your loop. Build up a query (not running) by adding {field: newValue} to $set clause. After the loop is done you will have an analog of {$set:{"a": 1, "b": 1, "c": 3}}. Then you will update all fields in one shot.
Here is official documentation:
http://docs.mongodb.org/manual/reference/operator/update/set/
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.
I have collections with huge amount of Documents on which I need to do custom search with various different queries.
Each Document have boolean property. Let's call it "isInTop".
I need to show Documents which have this property first in all queries.
Yes. I can easy do sort in this field like:
.sort( { isInTop: -1 } );
And create proper index with field "isInTop" as last field in it. But this will be work slowly, as indexes in mongo works best with unique fields.
So is there is solution to show Documents with field "isInTop" on top of each query?
I see two solutions here.
First: set Documents wich need to be in top the _id from "future". As you know, ObjectId contains timestamp. So I can create ObjectId with timestamp from future and use natural order
Second: create separate collection for Ducuments wich need to be in top. And do queries in it first.
Is there is any other solutions for this problem? Which will work fater?
UPDATE
I have done this issue with sorting on custom field which represent rank.
Using the _id field trick you mention has the problem that at some point in time you will reach the special time, and you can't change the _id field (without inserting a new document and removing the old one).
Creating a special collection which just holds the ones you care about is probably the best option. It gives you the ability to logically (and to some extent, physically) separate the documents.
Newly introduced in mongodb there is also support for a "sparse" index which may fulfill your needs as well. You could only set the "isInTop" field when you want it to be special, and then create a sparse index on it which would not have the problems you would normally have with a single indexed boolean field (in btrees).