All!
My document has such structure:
{
fname: value,
lname: value,
city: value
}
When I use find() method, I get result in default order fname, lname, city.
But I want to get result in other order of field, such as: city, fname, lname.
Does mongodb allow fields ordering in result?
Yes and no. To really do this sort of manipulation you need the aggregation framework. Even then it's a bit tricky since you must "rename" the fields in order to change their order. This is because there are general optimizations in place that "copy" content from one "pipeline" stage to another. This is considered optimal.
But you can always do this:
db.collection.aggregate([
{ "$project": {
"tcity": "$city",
"tfname": "$fname",
"tlname": "$lanme"
}},
{ "$project": {
"city": "$tcity",
"fname": "$tfname",
"lname": "$tlname"
}}
])
Very contrived, but that is the way you have to do it. Or otherwise just live with a single projection for the "renamed" fields in the order you want and then just "rename" again in code.
Or of course, your code can simply "re-map" the field names from the result of a query.
But the basic point is that MongoDB itself "preserves" the original order of fields as an "optimization" to how they are stored and does not mess with the output otherwise. If you want to you can, but you need to take the steps as shown in order to do so.
You can use .sort(). eg:
db.stocks.find().sort( { ticker: 1, date: -1 } )
For more info see http://docs.mongodb.org/manual/reference/method/cursor.sort/
I assume you are referring to the output as it shows up in the Mongo shell? If so, you can assign the returned output to a variable (cursor, I believe) and step through your cursor afterwards while outputting the individual field variables in any order you like.
Related
I don't want to expose the full object ID to the client, instead I want to show him only a short of the last 4 chars of the actual object ID of an entity in the collection.
For example: ObjectId("5fcca5d997239a74da0d67a9") will become just 67a9
So it will be much easier to "talk" with ids of documents instead of the full object it.
Then I need to find the document in the DB using only the 67a9.
Is this possible and how?
According to this issue in Jira the resolution is "Won't fix".
ObjectId is not a String, is another object, so $regex is no possible.
Check this example where $regex works ok when _id is an String but not an ObjectId.
So one possible option is duplicate every field _id in another field called id or whatever where the id is in String format.
Then, you can do this query:
db.collection.find({
"_id": {
"$regex": "67a9$"
}
})
Example here where I've added more _id fields that not match the pattern
As pointed out, regex won't work on an ObjectId. But there is an easy workaround. Just use aggregation to first convert your ObjectId into a string and then match it.
db.collection.aggregate([
{
$addFields: {
tempId: { $toString: '$_id' },
}
},
{
$match: {
tempId: { $regex: "67a9"}
}
}
])
Obviously not a great solution to use on very large collections.
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?
Let's say I have four documents in my collection:
{u'a': {u'time': 3}}
{u'a': {u'time': 5}}
{u'b': {u'time': 4}}
{u'b': {u'time': 2}}
Is it possible to sort them by the field 'time' which is common in both 'a' and 'b' documents?
Thank you
No, you should put your data into a common format so you can sort it on a common field. It can still be nested if you want but it would need to have the same path.
You can use use aggregation and the following code has been tested.
db.test.aggregate({
$project: {
time: {
"$cond": [{
"$gt": ["$a.time", null]
}, "$a.time", "$b.time"]
}
}
}, {
$sort: {
time: -1
}
});
Or if you also want the original fields returned back: gist
Alternatively you can sort once you get the result back, using a customized compare function ( not tested,for illustration purpose only)
db.eval(function() {
return db.mycollection.find().toArray().sort( function(doc1, doc2) {
var time1 = doc1.a? doc1.a.time:doc1.b.time,
time2 = doc2.a?doc2.a.time:doc2.b.time;
return time1 -time2;
})
});
You can, using the aggregation framework.
The trick here is to $project a common field to all the documents so that the $sort stage can use the value in that field to sort the documents.
The $ifNull operator can be used to check if a.time exists, it
does, then the record will be sorted by that value else, by b.time.
code:
db.t.aggregate([
{$project:{"a":1,"b":1,
"sortBy":{$ifNull:["$a.time","$b.time"]}}},
{$sort:{"sortBy":-1}},
{$project:{"a":1,"b":1}}
])
consequences of this approach:
The aggregation pipeline won't be covered by any of the index you
create.
The performance will be very poor for very large data sets.
What you could ideally do is to ask the source system that is sending you the data to standardize its format, something like:
{"a":1,"time":5}
{"b":1,"time":4}
That way your query can make use of the index if you create one on the time field.
db.t.ensureIndex({"time":-1});
code:
db.t.find({}).sort({"time":-1});
I have a mongoDB document that has the following structure:
{
user:user_name,
streams:[
{user:user_a, name:name_a},
{user:user_b, name:name_b},
{user:user_c, name:name_c}
]
}
I want to use $pullAll to remove from the streams array, passing it an array of streams (the size of the array varies from 1 to N):
var streamsA = [{user:"user_a", name:"name_a"},{user:"user_b", name:"name_b"}]
var streamsB = [{name:"name_a", user:"user_a"},{name:"name_b", user:"user_b"}]
I use the following mongoDB command to perform the update operation:
db.streams.update({name:"user_name", {"$pullAll:{streams:streamsA}})
db.streams.update({name:"user_name", {"$pullAll:{streams:streamsB}})
Removing streamsA succeeds, whereas removing streamsB fails. After digging through the mongoDB manuals, I saw that the order of fields in streamsA and streamsB records has to match the order of fields in the database. For streamsB the order does not match, that's why it was not removed.
I can reorder the streams to the database document order prior to performing an update operation, but is there an easier and cleaner way to do this? Is there some flag that can be set to update and/or pullAll to ignore the order?
Thank You,
Gary
The $pullAll operator is really a "special case" that was mostly intended for single "scalar" array elements and not for sub-documents in the way you are using it.
Instead use $pull which will inspect each element and use an $or condition for the document lists:
db.streams.update(
{ "user": "user_name" },
{ "$pull": { "streams": { "$or": streamsB } }}
)
That way it does not matter which order the fields are in or indeed look for an "exact match" as the current $pullAll operation is actually doing.
I have the following document structure:
{
_id: ...,
name: "Item1",
Props: [
{
Key: "numberKey",
Val: 1234
},
{
Key: "dateKey",
Val: Date("2013-09-09")
}]
}
This is simplified and there can be various Keys and Values in Props field in the real application.
My question - is it possible to $group and $sum "numberKey"s by "dateKey"s?
What structure should I use if this is not possible? I need users to let add keys and values so I need something flexible.
Unfortunately, that isn't possible using aggregation with your schema. The problem is that aggregation is meant to operate over values in an array that are being selected by the $group clause and those elements have all the data needed. Your setup separates what you want to group by and what you want to sum. You could use a mapReduce job to do what you want with your schema. http://docs.mongodb.org/manual/core/map-reduce/ should be able to get you started.
Let me know if you have any other questions.
Best,
Charlie