I have a mongoDB collection which contains a list of documents which contain (among other items) a time stamp and an address for each different type of device. The device address is in hexidecimal (eg. 0x1001, 0x2001, 0x3001, etc.).
On the server side, I'm trying to query the collection to see what documents exists within a certain date range, and for a list of device
Collection.find(
{"addr": data.devices.Controls, "time": {$gte:d0, $lte:d1}},{},
function (err, docs) {
if( err|| !docs) console.log("No data found");
else {
//I've simplified the code here...
}
}
);
d0 and d1 are my start and end dates... and the data.devices.Controls is a list of device addresses. If I add the line:
console.log("Controls: " + JSON.stringify(data.devices.Controls));
I can see on the server side that it prints out a list of addresses that I'm looking for (the actual print statement looks like: Controls: ["0x1001", "0x2001", "0x3001"].)
However, this find statement doesn't seem to return any data from the query. There's no error (as I don't see the "No Data Found" message)... It just doesn't seem to return any data. What's strange is that if I specify a specific element out of the Controls array (something like data.devices.Controls[0]...), then it works fine. I can specify any element in the array and it works... but by passing an entire array in the argument, it doesn't seem to work. Does anyone know why this happens (and how to fix it)?
You need to use the $in operator to match against an array of values; like this:
Collection.find(
{"addr": {$in: data.devices.Controls}, "time": {$gte:d0, $lte:d1}}, ...
Related
Although I think it is a general question, I could not find a solution that matches my needs.
I have 2 Mongo collections. The 'users' collection and the second one 'dbInfos'.
Now, I have a template called 'Infos' and want the already existing fields in the Mongo collections to be presented to the user in input fields in case there is data in the collection. When no data is provided in the database yet, it should be empty.
So here is my code, which works fine until I want to capture the fields from the second collection.
Template.Infos.onRendered(function() {
$('#txtName').val(Meteor.user().profile.name);
$('#txtEmail').val(Meteor.user().emails[0].address);
});
These 2 work great.
But I don´t know how to query the infos from the collection 'dbInfos', which is not the 'users' collection. Obviously Meteor.user().country does not work, because it is not in the 'users' collection. Maybe a find({}) query? However, I don´t know how to write it.
$('#txtCountry').val( ***query function***);
Regarding the structure of 'dbInfos': Every object has an _id which is equal to the userId plus more fields like country, city etc...
{
"_id": "12345",
"country": "countryX",
"city": "cityY"
}
Additionally, how can I guarantee that nothing is presented, when the field in the collection is empty? Or is this automatic, because it will just return an empty field?
Edit
I now tried this:
dbInfos.find({},{'country': 1, '_id': 0})
I think this is the correct syntax to retrieve the country field and suppress the output of the _id field. But I only get [object Object] as a return.
you're missing the idea of a foreign key. each item in a collection needs a unique key, assigned by mongo (usually). so the key of your country info being the same as the userId is not correct, but you're close. instead, you can reference the userId like this:
{
"_id": "abc123",
"userId": "12345",
"country": "countryX",
"city": "cityY"
}
here, "abc123" is unique to that collection and assigned by mongo, and "12345" is the _id of some record in Meteor.users.
so you can find it like this (this would be on the client, and you would have already subscribed to DBInfos collection):
let userId = Meteor.userId();
let matchingInfos = DBInfos.find({userId: userId});
the first userId is the name of the field in the collection, the second is the local variable that came from the logged in user.
update:
ok, i think i see where you're getting tripped it. there's a difference between find() and findOne().
find() returns a cursor, and that might be where you're getting your [object object]. findOne() returns an actual object.
for both, the first argument is a filter, and the second argument is an options field. e.g.
let cursor = DBInfos.find({
userId: Meteor.userId()
},
{
fields: {
country: 1
}
});
this is going to:
find all records that belong to the logged in user
make only the country and _id fields available
make that data available in the form of a cursor
the cursor allows you to iterate over the results, but it is not a JSON object of your results. a cursor is handy if you want to use "{{#each}}" in the HTML, for example.
if you simply change the find() to a findOne():
let result = DBInfos.findOne({ /** and the rest **/
... now you actually have a JSON result object.
you can also do a combination of find/fetch, which works like a findOne():
let result = DBInfos.find({
userId: Meteor.userId()
},
{
fields: {
country: 1
}
}).fetch();
with that result, you can now get country:
let country = result.country;
btw, you don't need to use the options to get country. i've been assuming all this code is on the client (might be a bad assumption). so this will work to get the country as well:
let result = DBInfos.findOne({userId: Meteor.userId()});
let country = result.country;
what's going on here? it's just like above, but the result JSON might have more fields in it than just country and _id. (it depends on what was published).
i'll typically use the options field when doing a find() on the server, to limit what's being published to the client. on the client, if you just need to grab the country field, you don't really need to specify the options in that way.
in that options, you can also do things like sort the results. that can be handy on the client when you're going to iterate on a cursor and you want the results displayed in a certain order.
does all that make sense? is that what was tripping you up?
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.
})
A mongo db has documents that look like:
{
"_id": : ObjectId("55cb43e8c78b04f43f2eb503"),
<some fields>
"topics": {
"test/23/result": 149823788,
"test/27/result": 147862733,
"input/misc/test": 14672882
}
}
I need to find all documents that have a topics field that contains a particular key. i.e. find all documents that have a topics.key = "test/27/result"
I've tried a number of things but none work yet, neither attempt below work,
they return no records event though some should match:
db.collName.find({"topics.test/27/result": {$exists:true}});
db.collName.find({"topics.test\/27\/result": {$exists:true}});
How can I make the query work?
The slash characters are inserted by another process. They are mqtt topic names.
I found the solution to my problem:
I was building the query wrong in my code. In the example below, evtData.source contains the key name to search for, i.e. 'test/27/result'
The query methodology that works for me is:
var query = {};
query['topics.' + evtData.source] = {$exists: true};
db.collName.find(query)
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.