I am trying to update an array element inside a document without changing whole array.
Array elements look like this
Suppose i have to only update index 1 element's value. For that i have:
_id of the document
index 1's value ("optionalImages-624476a7bd4d2bfe6bf86e9a-1-1650025533684.jpeg")
to be updated value ("optionalImages-624476a7bd4d2bfe6bf86e9a-1-1650025534589.jpeg").
I think it can be updated by mongodb's arrayfilters but i don't get the documentation correctly.
Your help will be highly appreciated.
Query1
arrayFilters using $[m] inside the path to specify the member value that we want to change
m is the member with value 20, and we set it to 100
(instead of 20 and 100, put your "....jpg" strings)
Playmongo
update(
{"_id": {"$eq": 1}},
{"$set": {"ar.$[m]": 100}},
{"arrayFilters": [{"m": {"$eq": 20}}])
Query2
pipeline update >= MongoDB 4.2
uses map on the array to
find the member with value 20, and replaces it with 100
Playmongo
update(
{"_id": {"$eq": 1}},
[{"$set":
{"ar":
{"$map":
{"input": "$ar",
"in": {"$cond": [{"$eq": ["$$this", 20]}, 100, "$$this"]}}}}}])
Related
I have a database with a collection of about 90k documents. Each document is as follows:
{
'my_field_name': "a", # Or "b" or "c" ...
'p1': Array[30],
'p2': Array[10000]
}
There are about 9 unique values for a field name. When there where ~30k documents in the collection:
>>> db.collection.distinct("my_field_name")
["a", "b", "c"]
However, now with 90k documents, db.collection.distinct() returns an empty list.
>>> db.collection.distinct("my_field_name")
[]
Is there a maxTimeMS setting for db.collection.distinct? If so how could I set it to a higher value. If not what else could I investigate?
One thing you can do to immediately speed up your query's execution time is to index the field on which you are running the 'distinct' operation on (if the field is not already indexed).
That being said, if you want to set a maxTimeMS, one work around is to rewrite your query as an aggregation and set the operation timeout on the returned cursor. E.g:
db.collection.aggregate([
{ $group: { _id: '$my_field_name' } },
]).maxTimeMS(10000);
However unlike distinct, a cursor will be returned by the above query.
I have a couchbase document as
{
"last": 123,
"data": [
[0, 1.1],
[1, 2.3]
]
}
currently have code to upsert the document to change the last property and add values to the data array, however, cannot find a way to insert unique values only. I'd like to avoid fetching the whole document and doing the filtering in javascript. Is there any way in couchbase?
arrayAddUnique will fail, cause there are floats in the subarrays per couchbase docs.
.mutateIn(`document`)
.upsert("last", 234)
.arrayAppend("data", newDataArray)
.execute( ... )
I have this aggregate query :
cr = db.last_response.aggregate([
{"$unwind": '$blocks'},
{"$match": {"sender_id": "1234", "page_id": "563921", "blocks.tag": "pay1"}},
{"$project": {"block": "$blocks.block"}}
])
Now i want to get the number of element it returned (is it empty cursor or not).
This is how i did :
I defined an empty array :
x = []
I iterated through the cursor and append the array x:
for i in cr :
x.append(i['block'])
print("length of the result of aggregation cursor :",len(x))
My question is : Is there any faster way to get the number of the result of aggregate query like the count() method of the find() query ?
Thanks
The faster way is that reject operations of transfers all data from mongod to you application. To do this you may add final group stage to count docs
{"$group": {"_id": None, "count": {"$sum": 1}}},
This is mean that mongod do aggregate and get as result count of docs.
Thereis no way to get count of result without execution of aggregation pipeline.
I have an array lets say [1,2,3] and a collection called 'Numbers' and it has a field called 'value'. I need to retain all the values in the array which are present against the 'value' field in any document in the collection.
Example,
Test array - [1,2,3]
Numbers collection - [{value: 1}, {value: 3}]
Result should be - [1,3]
Result is that way because '2' was not present against 'value' field in any documents within 'Numbers' collection.
How do i do this?
You can try below distinct query with projection and query filter.
db.Numbers.distinct( "value", { "value": { $in: [1,2,3] } } )
I have a MongoDB document with quite a large embedded array:
name : "my-dataset"
data : [
{country : "A", province: "B", year : 1990, value: 200}
... 150 000 more
]
Let us say I want to return data objects where country == "A".
What is the proper way of doing this, for example via NodeJs?
Given 150 000 entries with 200 matches, how long should the query take approximately?
Would it be better (performance/structure wise) to store data as documents and the name as a property of each document?
Would it be more efficient to use Mysql for this? )
A) Just find them with a query.
B) If the compound index {name:1, data.country:1} is built, the query should be fast. But you store all the data in one array, $unwind op has to be used. As a result, the query could be slow.
C) It will be better. If you store the data like:
{country : "A", province: "B", year : 1990, value: 200, name:"my-dataset"}
{country : "B", province: "B", year : 1990, value: 200, name:"my-dataset"}
...
With compound index {name:1, country:1}, the query time should be < 10ms.
D) MySQL vs MongoDB 1000 reads
1.You can use the MongoDB aggregation :
db.collection.aggregate([
{$match: {name: "my-dataset"}},
{$unwind: "$data"},
{$match: {"data.country": "A"}}
])
Will return a document for each data entry where the country is "A". If you want to regroup the datasets, add a $group stage :
db.collection.aggregate([
{$match: {name: "my-dataset"}},
{$unwind: "$data"},
{$match: {"data.country": "A"}},
{$group: {_id: "$_id", data: {$addToSet: "$data"}}}
])
(Didn't test it on a proper dataset, so it might be bugged)
2.150000 Subdocuments is still not a lot for mongodb, so if you're only querying on one dataset it should be pretty fast (the order of the millisecond).
3.As long as you are sure that your document is going to be smaller than 16MB (kinda hard to say), the maximum BSON document size), it should be fine, but the queries would be simpler if you stored your data as documents with the dataset name as a property, which is generally better for performances.