I have a collection of simple documents like:
{tag: "...", data: {...}}
What i want to do is to find a document with given tag and make a projection from its data. But if this document cannot be found, i want to aggregate data from another documents.
I wonder if there is a way to make it using single operation.
Related
Example :[{
inst:"EVA",
std:"12th"
},
{
inst:"KSF",
std:"12th"
}]
As per the above example, In my case if "inst: "EVA" is already there in the "qualification" array so we need to update the object from the existing one.
Then "inst: "KSF" does not already exist in the "qualification" array so we need to add that one.
Help me if there is any way to upsert using firestore query.
There is no "upsert" operation for objects in arrays. If you need to make changes to that array, you will have to read the document, modify the contents of the array in memory, then update the document with the new contents of the array.
Arrays of objects usually do not work the way that people want, given their limitations on querying and updating. It's usually better to store data as documents in a nested subcollection, so they can be more easily queried and updated by the contents of their fields.
I'm using mongodb for a project. Need to know which would be a better implementation for queries.
Consider I have to search for 10 documents out of a total 1000 documents based on a condition (not id).
Would it be better to query using document _id's (after storing the required id's in another collection beforehand by checking for the condition whenever insertion is done)
OR
Would it better to traverse all the documents and get the required documents using the condition
The main aim here is to split documents into different categories and display the documents belonging to a particular category. So storing id's of documents belonging to each category or search for documents in that category by traversing through all the documents?
I have heard that mongodb uses hashed indexing (so feel option 1 would be faster), but I couldnt find anything regarding that. So a small description regarding document storage and queries would also be good.
The optimum way to query for the cuisine type example would be to store what the restaurant serves in an array of strings or objects, and index that field.
For example:
{
name: "International House"
cuisine: [
{ name: "Chinese", subtype: "Kowloon"},
{ name: "Japanese", subtype: "Yakitori"},
{ name: "American", subtype: "TexMex" }
]
}
Then create an index on { "cuisine.name": 1 }.
When you need to find all restaurants that serve Chinese food, the query:
db.collection.find({"cuisine.name":"Chinese")
will use that index, and only scan the documents that match.
If I have a data with a structure like this as a single document in a collection:
{
_id: ObjectId("firstid"),
"name": "sublimetest",
"child": {
_id: ObjectId("childid"),
"name": "materialtheme"
}
}
is there a way to search for the embedded document by the id "childid" ?
because mongo doesn't index the _id fields of embedded documents (correct me if I am wrong here),
as this query doesn't work :
db.collection.find({_id:"childid"});
Also please suggest me if there is any other document database that would be suitable for this kind of retreiving data that is structured as a tree, where the requirement is to :
query children without having to issue joins
find any node in the tree as fast as you would find the root node, as if all these nodes were stored as separate documents in a collection.
Why this is not a duplicate of question(s) suggested :
the potential-duplicate-question, queries document by using dot notation. But what if the document is nested 7 levels deep ? In such case it would not be suitable to write a query using dot notation. what I want is that, all documents, whether top level, or nested, if they have the _id field, should be in the bucket of _id indexes, so that when you search db.collection.find({_id: "asdf"}), it should take into account documents that are nested too that have the _id field matching "asdf". In short, it should be as if the inner document weren't nested, but present parallel to the outer one.
You can use the dot notation:
db.posts.find({"child._id": "childid"})
I want to remove all documents with the matched query in mongodb. which means there will be a field "head" in all collections. i want to remove all documents in each collection with head is matched with id : 128643 using single query. How can i do that with mongoose?
I’d recommend spending time in the mongoose documentation, it’s pretty easy to find there...
The command you’re looking for is Model.deleteMany()
So in your case, it would be Model.deleteMany({ id: 128643 });
Say I have objects in a MongoDB collection:
{
...
"json" : "{\"things\":[2494090781803658355,5114030115038563045,3035856943768375362,8931213615561493991,7574631742057150605,480863244020297489]}"
}
It's an Azure "MongoDB" so doesn't support all the features, but suppose it does.
This search will find that document:
db.coll.find({"json" : {$regex : "5114030115038563045|8931213615561493991"}})
Of course, it's scanning the whole collection to pull these records out. What's an efficient/faster way to find documents where the list of "things"
contains any of a list of "things" in a query? It seems like throwing a search engine like Solr or ElasticSearch would solve this, and perhaps
using another Azure's Data Lake storage would make this more searchable, so I'm considering those options. They're outside the scope of this
question though; I'd like to know if there's a Mongo-ish way to search this collection by index.
The only option you have available to you if you're storing a JSON string is to use a text index with a $text operator.
If this document structure isn't set in stone, however, you might consider also separately storing the JSON as a nested subdocument (with the appropriate sanitation, of course). This would allow you to construct an index on json.things, while still storing the JSON string, and allow you to perform a query on e.g. "json.things": {$in: [ "5114030115038563045", "8931213615561493991" ]}