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.
Related
enter image description here
Let's say that we have many documents like this in the photo
I have the above schema. I want to find the document based on _id first and then push an array of values to providedServices which belongs to the _id which is inside barbers array
A little help. Can't seem to find this out!
You need to find the related arrays firstly. For this, you can use $elemMatch or write it as 'barbers._id' : {$elemMatch: parameter}' .
Here we tried to find document with filtering it's own id and barbers id. You can change the filter as you wished. It can be only search on barbers id.
Need to write your DocumentName and your parameters instead of idValue, barbersId, serviceModel.
const result = await DocumentName.findOneAndUpdate(
{
$and:
[
{_id: mongoose.Types.ObjectId(idValue)},
{'barbers': {$elemMatch: {_id: mongoose.Types.ObjectId(barbersId)}}}
]
},
{ $push: { 'barbers.$.providedServices': serviceModel } },
{ new: true })
At first, we found the related barbers array inside of all documents. Then we pushed the model inside of providedServices array into this barbers array.
I am using mongo server version 3.4, so my question pertains to the functionality of that version. I cannot upgrade anytime soon, so please keep that in mind. If have a field in some documents in a MongoDB collection that may contain a string but also have trailing characters, how might I find them when submitting multiple "startsWith" strings to be evaluated in the same query? I may have some difficulty explaining this, so let me show some examples. Let's say that I have a field called "description" in all of my documents. This description might be encoded so that the text is not completely straightforward. Some values might be:
green:A-4_ABC
yellow:C-12_456
red:A-431_ZXCVQ
yellow_green:C-12_999
brown:B-3_R
gray:EN-44_195
EDIT: I think I made a mistake with using words in my keys. The keys are a randomized string of numbers, letters, and underscores, followed by a colon, then one to three letters, followed by a dash, then a couple of numbers, then an underscore, and lastly followed by several alphanumeric characters:
LKEF543SLI54EH2J897FQ_HF234EWOH:ZX-82_FR2
I realize that this sounds arbitrary and stupid, but it is an encoding of information that is intended to result in a unique key. It is in data that I receive, so I cannot change it, unfortunately.
Now, I want to find all of the documents with descriptions that start with any of the following values, and all of these values must be submitted in the same query. I might have hundreds of submitted values, and I need to get all matching documents at once. Here is a short list of what might be submitted in a single query:
green:A-4
red:A-431
gray:EN-44
yellow_green:C-12
Note that it was not accidental that the text is everything prior to the last underscore. And, as with one of the examples, there might be more than one underscore. With my use case, I cannot create a query that hard-codes these strings in the javascript regex format. And the $in filter does not work with "startsWith" functionality, particularly when you pass in a list of strings (though I am familiar with supplying a list of hard-coded javascript regexes). Is there any way to use the $in operator where I can take a list of strings that are passed in from the user who wants to run a query like this? Or is there something equivalent? The cherry on the top of all of this would be to find a way to project the matching document with the string that it matched (either from the query, or by some substring magic that I cannot seem to figure out).
EDIT: Specifically, when I find each document, I want to be able to project everything from they key up until the LAST underscore, like:
LKEF543SLI54EH2J897FQ_HF234EWOH:ZX-82
(along with its value)
Thanks in advance for any nudges in the right direction.
We use $objectToArray to get {k:field_name, v:field_value} array. Then we split by _ token all values and convert to object with $arrayToObject operator.
Next step we apply $match operator to filter documents and exclude data with $unset.
Note: If your document contains array or subdocuments, we may use $filter before we convert $objectToArray.
db.collection.aggregate([
{
$addFields: {
data: {
$arrayToObject: {
$map: {
input: {
$objectToArray: "$$ROOT"
},
in: {
k: "$$this.k",
v: {
$arrayElemAt: [
{
$split: [
{
$toString: "$$this.v"
},
"_"
]
},
0
]
}
}
}
}
}
}
},
{
$match: {
"data.green": "A-4",
"data.red": "A-431",
"data.gray": "EN-44",
"data.yellow_green": "C-12"
}
},
{
$unset: "data"
}
])
MongoPlayground
This question already has answers here:
MongoDB - get documents with max attribute per group in a collection
(2 answers)
Closed 5 years ago.
I have a collection:
{'_id':'008','name':'ada','update':'1504501629','star':3.6,'desc':'ok', ...}
{'_id':'007','name':'bob','update':'1504501614','star':4.2,'desc':'gb', ...}
{'_id':'005','name':'ada','update':'1504501532','star':3.2,'desc':'ok', ...}
{'_id':'003','name':'bob','update':'1504501431','star':4.5,'desc':'bg', ...}
{'_id':'002','name':'ada','update':'1504501378','star':3.4,'desc':'no', ...}
{'_id':'001','name':'ada','update':'1504501325','star':3.6,'desc':'ok', ...}
{'_id':'000','name':'bob','update':'1504501268','star':4.3,'desc':'gg', ...}
...
if I want the result is, the max value of 'update' of the same 'name', means the newest document of 'name', get the whole document:
{'_id':'008','name':'ada','update':'1504501629','star':3.6,'desc':'ok', ...}
{'_id':'007','name':'bob','update':'1504501614','star':4.2,'desc':'gb', ...}
...
How to do it most effective?
I do it now in python is:
result=[]
for name in db.collection.distinct('name'):
result.append(db.collection.find({'name':name}).sort('update',-1)[0])
is it do 'find' too many times?
=====
I do this for crawl data with 'name', get many other keys, and every time I insert a document, I set a key named 'update'.
When I using the database, I want the newest document of specific 'name'. so it looks can not just use $group.
How should I do? re design the db structure or better way to find?
=====
Improved !
I've tried create index of 'name' & 'update', the process is shortened from half hour to 30 seconds!
But I still welcome for better solution ^_^
Your use case scenario suits real good for aggregation. As I see in your question you already know that but can't figure out how to use $group and take whole document that has the max update. If you $sort your documents before $groupyou can use $firstoperator. So no need to send a find query for each name.
db.collection.aggregate(
{ $sort: { "name": 1, "update": -1 } },
{ $group: { _id: "$name", "update": { $first: "$update" }, "doc_id": { $first: "$_id" } } }
)
I did not add an extra $projectoperation to aggregate, you can just add fields that you want in result to $groupwith $firstoperator.
Additionally, if you look closer to $sortoperation, you can see it uses your newly created index, so you did good to add that, otherwise I will recommend it too :)
Update: For your question in comment:
You should write all keys in $group. But if you think it will look bad or new fileds will come in future and does not want to rewrite $groupeach time, I would do that:
First get all _idfields of desired documents in aggregation and then get these documents in one findquery with $inoperator.
db.collection.find( { "_id": { $in: [<ids returned in aggregation] } } )
I am using Mongo 3.2.14
I have a mongo collection that looks like this:
{
'_id':...
'field1':...
'field2':...
'field3':...
etc...
}
I want to aggregate this way:
db.collection.aggregate{
'$match':{},
'$project':{
'field1':1,
'field2':1,
'field3':1,
etc...(all fields)
}
}
Is there a way to include all fields in the project without listing each field one by one ? (I have around 30 fields, and growing...)
I have found info on this here:
MongoDB $project: Retain previous pipeline fields
Include all existing fields and add new fields to document
how to not write every field one by one in project
However, I'm using mongo 3.2.14 and I don't need to create a new field, so, I think I cannot use $addFields. But, if I could, can someone show me how to use it?
Basically, if you want all attributes of your documents to be passed to the next pipeline you can skip the $project pipeline. but if you want all the attributes except the "_id" value then you can pass
{ $project: { _id: 0 } }
which will return every value except the _id.
And if by any chance you have embedded lists or nests that you want to flatten, you can use the $unwind pipeline
you can use $replaceRoot
db.collection.aggregate{
"$match": {},
"$project": {
"document": "$$ROOT"
},
{
"$replaceRoot": { "newRoot": "$document" }
}
this way you can get the exact document returned with all the fields in the document...you don't need to add each field one by one in the $project...try using $replaceRoot at the end of the pipeline.
So in my MongoDB test the _id field is a unique string in format of "SensorSN:YrMonthDay"
What I want to query is all data for 2016 month 3. But I can't quite get the Regex right.
So here is some sample _ids:
12345678:2016325
87654321:2016325
Basically I want to return both of the documents because they are in the month of March.
My first attempts have been similar to db.collection.find({'_id':'/:20163/'}) and so far no luck.
Thanks!
Take off the quotes, the find is looking for a string matching exactly that instead of doing the regex search.
db.collection.find({ '_id': /:20163/ })
or
db.collection.find({ _id: /:20163/ })
both work fine for me!
Your case would only match if your sample document was:
{ _id: "/:20163/" }
Remove the quotes
db.collection.find({ _id: /:20163/ });
Or try using regex -
db.collection.find({ _id: { $regex: /:20163/ } });
https://docs.mongodb.com/manual/reference/operator/query/regex/