Given a CouchDb view that emits like this:
emit([doc.name, doc.date], doc)
How to filter on multiple doc.name values? Below doesn't seem to work.
keys=[["name1",{}],["name2",{}],["name3",{}]]
When you specify the keys parameter, you are specifying an exact match. Assuming you want a range query (every document with name="name1" no matter the date) then you need to query with a startkey and an endkey:
?startkey=["name1", ""]&endkey=["name1", {}]
And will have to repeat the query for each name.
Related
I have been trying to figure out a way to query a list of documents where I have a range filter on one field and order by another field which of course isn't possible, see my other question: Order by timestamp with range filter on different field Swift Firestore
But is it possible to save documents with the timestamp as id and then it would sort by default? Or maybe hardcode an ID, then retrieve the last created document id and increase id by one for the next post to be uploaded?
This shows how the documents is ordered in the collection
Any ideas how to store documents so they are ordered by created at in the collection?
It will order by document ID (ascending) by default in Swift.
You can use .order(by: '__id__') but the better/documented way is with FieldPath documentID() I don't really know Swift but I assume that it's something like...
.order(by: FirebaseFirestore.FieldPath.documentID())
JavaScript too has an internal variable which simply returns __id__.
.orderBy(firebase.firestore.FieldPath.documentId())
Interestingly enough __name__ also works, but that sorts the whole path, including the collection name (and also the id of course).
If I correctly understood your need, by doing the following you should get the correct order:
For each document, add a specific field of type number, called for example sortNbr and assign as value a timestamp you calculate (e.g. the epoch time, see Get Unix Epoch Time in Swift)
Then build a query sorted on this field value, like:
let docRef = db.collection("xxxx")
docRef.order(by: "sortNbr")
See the doc here: https://firebase.google.com/docs/firestore/query-data/order-limit-data
Yes, you can do this.
By default, a query retrieves all documents that satisfy the query in
ascending order by document ID.
See the docs here: https://firebase.google.com/docs/firestore/query-data/order-limit-data
So if you find a way to use a timestamp or other primary key value where the ascending lexicographical ordering is what you want, you can filter by any fields and still have the results sorted by the primary key, ascending.
Be careful to zero-pad your numbers to the maximum precision if using a numeric key like seconds since epoch or an integer sequence. 10 is lexicographical less than 2, but 10 is greater than 02.
Using ISO formatted YYYY-mm-ddTHH:MM:SS date-time strings would work, because they sort naturally in ascending order.
The order of the documents shown in the Firebase console is mostly irrelevant to the functioning of your code that uses Firestore. The console is just for browsing data, and that sorting scheme makes it relatively intuitive to find a document you might be looking for, if you know its ID. You can't change this sort order in the console.
Your code is obviously going to have other requirements, and those requirements should be coded into your queries, without regarding any sort order you see in the dashboard. If you want time-based ordering of your documents, you'll have to store some sort of timestamp field in the document, and use that for ordering. I don't recommend using the timestamp as the ID of a document, as that could cause problems for you in the future.
I have two type of documents in a mongodb collection:
one where key sessions has a simple value:
{"sessions": NumberLong("10000000000001")}
one where key sessions has an array of values.
{"sessions": [NumberLong("10000000000001")]}
Is there any way to retrieve all documents from the second category, ie. only documents whose value is an arary and not a simple value?
You can use this kind of query for that:
db.collectionName.find( { $where : "Array.isArray(this.sessions)" } );
but you'd better convert all the records to one type to keep the things consistent.
This code can be simple like this:
db.c.find({sessions:{$gte:[]}});
Explanation:
Because you only want to retrieve documents whose sessions data type is array, and by the feature of $gte (if data types are different between tow operands, it returns false; Double, Integer32, Integer64 are considered as same data type.), giving an empty array as the opposite operand will help to retrieve all results by required.
Also , $gt, $lt, $lte for standard query (attention: different behaviors to operaors with same name in expression of aggregation pipeline) have the same feature. I proved this by practice on MongoDB V2.4.8, V2.6.4.
I tried running this:
db.col.find().skip(5).distinct("field1")
But it throws an error.
How to use them together?
I can use aggregation but results are different:
db.col.aggregate([{$group:{_id:'$field1'}}, {$skip:3},{$sort:{"field1":1}}])
What I want is links in sorted order i.e numbers should come first then capital letters and then small letters.
Distinct method must be run on COLLECTION not on cursor and returns an array. Read this
http://docs.mongodb.org/manual/reference/method/db.collection.distinct/
So you can't use skip after distinct.
May be you should use this query
db.col.aggregate([{$group:{_id:'$field1'}}, {$skip:3},{$sort:{"_id":1}}]) because field field1 will not exists in result after first clause of grouping.
Also I think you should do sort at first and then skip because in your query you skip 3 unsorted results and then sort them.
(If you provide more information about structure of your documents and what output you want it would be more clearly and I will correct answer properly)
I'm trying to do create a query to exclude all documents which have an empty/null value in one specific field.
What is the query syntax or the programmatic way to do this?
You can use a required range query, which is open at both sides, like:
+field:[* TO *]
That is probably adequate, assuming that the documents to exclude have no value in the index.
If some form of default value appears, you would have to exclude that value as well, like:
+field:[* TO *] -field:NULL
What I'm trying to do:
Filter a field of a collection that matches a given condition. Instead of returning every item in the field (which is an array of items), I only want to see matched items.
Similar to
select items from test where items.histPrices=[10,12]
It is also similar to what's found on the mongodb website here: http://www.mongodb.org/display/DOCS/Retrieving+a+Subset+of+Fields
Here's what I have been trying:
db.test.save({"name":"record", "items":[{"histPrices":[10,12],"name":"stuff"}]})
db.test.save({"name":"record", "items":[{"histPrices":[10,12],"name":"stuff"},
{"histPrices":[12,13],"name":"stuff"},{"histPrices":[11,14],"name":"stuff"}]})
db.test.find({},{"name":1,"items.histPrices":[10, 12]})
It will return all the objects that have a match for items.histPrices:[10,12], including ALL of the items in items[]. But I don't want the ones that don't match the condition.
From the comments left on Mongodb two years ago, the solution to get only the items with that histPrices[10,12] is to do it with javascript code, namely, loop through the result set and filter out the other items.
I wonder if there's a way to do that with just the query.
Your find query is wrong
db.test.find({},{"name":1,"items.histPrices":[10, 12]})
Your condition statement should be in the first part of the find statement.In your query {} means fetch all documents similar to this sql
select items from test (no where clause)
you have to change your mongodb find to
db.test.find({"items.histPrices":[10, 12]},{"name":1})
make it work
since your items is an array and if you wanted to return only the matching sub item, you have to use positional operator
db.test.find({"items.histPrices":[10, 12]},{"name":1,'items.$':1})
When working with arrays Embedded to the Document, the best approach is the one suggested by Chien-Wei Huang.
I would just add another aggregation, with the $group (in cases the document is very long, you may not want to retrieve all its content, only the array elements) Operator.
Now the command would look like:
db.test.aggregate({$match:{name:"record"}},
{$unwind:"$items"},
{$match {"items.histPrices":[10, 12]}},
{$group: {_id: "$_id",items: {$push: "$items"}}});)
If you are interested to return only one element from the array in each collection, then you should use projection instead
The same kind of issue solved here:
MongoDB Retrieve a subset of an array in a collection by specifying two fields which should match
db.test.aggregate({$unwind:"$items"}, {$match:{"items.histPrices":[10, 12]}})
But I don't know whether the performance would be OK. You have to verify it with your data.
The usage of $unwind
If you want add some filter condition like name="record", just add another $march at first, ex:
db.test.aggregate({$match:{name:"record"}}, {$unwind:"$items"}, {$match:{"items.histPrices":[10, 12]}})
https://jira.mongodb.org/browse/SERVER-828
Get particular element from mongoDB array
MongoDB query to retrieve one array value by a value in the array