I have a collection which have a document like this:
collection 1
{
_id: ObjectID(),
name: foo
}
I would get ObjectID of the above collection and copy into a document of another collection in order to reference correctly. Should I do simply:
db.collection1.find({name:"foo"},{_id:1})
EDIT
A call to find will return a cursor. Cursors works like an iterator in other languages. You can either attempt to find the first element in the cursor using the next() function and then get it's _id property or simplify your statement using findOne:
var x = db.collection1.findOne({name:"foo"}, {_id:1});
var id = x._id;
This is making an assumption that you are getting a document back from that query. You'll probably want to add a null check on x before grabbing the _id property.
Related
Why dosen't db.find work? The console.log gets undefined...
var course = (db.courses.find({ _id: mongo.helper.toObjectID(param.course)}));
console.log(course.body)
The way you are trying use Selects documents in a collection and returns a cursor to the selected documents., so you can't use the way you are trying to use it.
You need to use a callback() to get the records matching the query.
The below code will give result in an array format :-
db.courses.findOne({ _id: mongo.helper.toObjectID(param.course)}).toArray(function(err, result)
{
console.log(result[0]); // will give you the matched record.
})
I'm having trouble updating a schema with a nested array to have one objectID in multiple rows of the array. Please help if you can!
My Schema looks like this:
var ProblemSchema = new Schema({
data : "etc..."
array1 : [{
_id : { type: 'String' },
array2: [{ type: Schema.Types.ObjectId}]
}]}
I can easily add an objectId to a SINGLE nested array using the $ operator in the update object like so:
var query = {array1._id : ID};
var update = {$push : {'array1.$.array2' : objectId}};
Problem.update(query, update)
I want to be able to, however, add the same objectId to MULTIPLE array2s (i.e. corresponding to different array1 ids) in one query, but the $ operator only matches to the first match to the query so I can't do it the way outlined above.
I'm also trying to avoid having a for loop on the server (I realize I could find() the correct document then use a for loop in which I manually push() the objectID to the correct array2s but the solution involves large overhead), can anyone think of a solution?
As of MongoDB 2.6, there's no way to do this with with a single update query. The $ positional operator is a placeholder for the first matching element in an array. There's a MongoDB feature request for updating all matching elements of an array, SERVER-1243, that you can track in the meantime.
In mongo I have a collections with records. These record are very complex. Now I would like to duplicate one of them.
I can easily select the one
mongo> var row = db.barfoo.find({"name":"bar"});
Now I actually don't know what to do. I don't know what is in row because I cannot find a way to print its content. How can I change specific properties and finally insert this modified row again
mongo> db.barfoo.insert(row);
thnx
You must change value _id - generate new:
var row = db.barfoo.findOne({"name":"bar"});
row._id = ObjectId();
db.barfoo.insert(row);
Good Luck!
I am going to assume that you're working directly inside the mongo shell.
Once you have your document (not a row :P ), you'd modify the properties in the same way you would a normal JavaScript object:
var doc = db.barfoo.findOne( { "name": "bar" } );
doc.name = "Mr Bar";
Note that the find() command returns a cursor, so if you're looking to extract a single document, you should use the findOne() function. This function returns a single document.
If you are interested in duplicating numerous documents, you can use the find() function and iterate over the cursor to retrieve each document:
db.barfoo.find( { "name": "bar" } ).forEach( function( doc ){
doc.name = "Mr Bar";
}
After you change the relevant properties, you can use the insert/save methods to persist the data back to mongo. Don't forget to change/delete the _id attribute so that you'll actually create a new document.
As a side note, in order to view the contents of an object in the mongo shell, you can use the print() function. If you want a more visually appealing output, you could use printjson().
I'm new to Mongo, and I'm trying to retrieve the first document from a find() query:
> db.scores.save({a: 99});
> var collection = db.scores.find();
[
{ "a" : 99, "_id" : { "$oid" : "51a91ff3cc93742c1607ce28" } }
]
> var document = collection[0];
JS Error: result is undefined
This is a little weird, since a collection looks a lot like an array. I'm aware of retrieving a single document using findOne(), but is it possible to pull one out of a collection?
The find method returns a cursor. This works like an iterator in the result set. If you have too many results and try to display them all in the screen, the shell will display only the first 20 and the cursor will now point to the 20th result of the result set. If you type it the next 20 results will be displayed and so on.
In your example I think that you have hidden from us one line in the shell.
This command
> var collection = db.scores.find();
will just assign the result to the collection variable and will not print anything in the screen. So, that makes me believe that you have also run:
> collection
Now, what is really happening. If you indeed have used the above command to display the content of the collection, then the cursor will have reached the end of the result set (since you have only one document in your collection) and it will automatically close. That's why you get back the error.
There is nothing wrong with your syntax. You can use it any time you want. Just make sure that your cursor is still open and has results. You can use the collection.hasNext() method for that.
Is that the Mongo shell? What version? When I try the commands you type, I don't get any extra output:
MongoDB shell version: 2.4.3
connecting to: test
> db.scores.save({a: 99});
> var collection = db.scores.find();
> var document = collection[0];
In the Mongo shell, find() returns a cursor, not an array. In the docs you can see the methods you can call on a cursor.
findOne() returns a single document and should work for what you're trying to accomplish.
So you can have several options.
Using Java as the language, but one option is to get a db cursor and iterate over the elements that are returned. Or just simply grab the first one and run.
DBCursor cursor = db.getCollection(COLLECTION_NAME).find();
List<DOCUMENT_TYPE> retVal = new ArrayList<DOCUMENT_TYPE>(cursor.count());
while (cursor.hasNext()) {
retVal.add(cursor.next());
}
return retVal;
If you're looking for a particular object within the document, you can write a query and search all the documents for it. You can use the findOne method or simply find and get a list of objects matching your query. See below:
DBObject query = new BasicDBObject();
query.put(SOME_ID, ID);
DBObject result = db.getCollection(COLLECTION_NAME).findOne(query) // for a single object
DBCursor cursor = db.getCollection(COLLECTION_NAME).find(query) // for a cursor of multiple objects
in mongodb records are store like this
{_id:100,type:"section",ancestry:nil,.....}
{_id:300,type:"section",ancestry:100,.....}
{_id:400,type:"problem",ancestry:100,.....}
{_id:500,type:"section",ancestry:100,.....}
{_id:600,type:"problem",ancestry:500,.....}
{_id:700,type:"section",ancestry:500,.....}
{_id:800,type:"problem",ancestry:100,.....}
i want to fetch records in order like this
first record whose ancestry is nil
then all record whose parent is first record we search and whose type is 'problem'
then all record whose parent is first record we search and whose type is 'section'
Expected output is
{_id:100,type:"section",ancestry:nil,.....}
{_id:400,type:"problem",ancestry:100,.....}
{_id:800,type:"problem",ancestry:100,.....}
{_id:300,type:"section",ancestry:100,.....}
{_id:500,type:"section",ancestry:100,.....}
{_id:600,type:"problem",ancestry:500,.....}
{_id:700,type:"section",ancestry:500,.....}
Try this MongoDB shell command:
db.collection.find().sort({ancestry:1, type: 1})
Different languages, where ordered dictionaries aren't available, may use a list of 2-tuples to the sort argument. Something like this (Python):
collection.find({}).sort([('ancestry', pymongo.ASCENDING), ('type', pymongo.ASCENDING)])
#vinipsmaker 's answer is good. However, it doesn't work properly if _ids are random numbers or there exist documents that aren't part of the tree structure. In that case, the following code would work rightly:
function getSortedItems() {
var sorted = [];
var ids = [ null ];
while (ids.length > 0) {
var cursor = db.Items.find({ ancestry: ids.shift() }).sort({ type: 1 });
while (cursor.hasNext()) {
var item = cursor.next();
ids.push(item._id);
sorted.push(item);
}
}
return sorted;
}
Note that this code is not fast because db.Items.find() will be executed n times, where n is the number of documents in the tree structure.
If the tree structure is huge or you will do the sort many times, you can optimize this by using $in operator in the query and sort the result on the client side.
In addition, creating index on the ancestry field will make the code quicker in either case.