how can I return a specific value for a specific document in MongoDB? For example, I have a schema that looks like:
{
"_id" : "XfCZSje7GjynvMZu7",
"createdAt" : ISODate("2015-03-23T14:52:44.084Z"),
"services" : {
"password" : {
"bcrypt" : "$2a$10$tcb01VbDMVhH03mbRdKYL.79FPj/fFMP62BDpcvpoTfF3LPgjHJoq"
},
"resume" : {
"loginTokens" : [ ]
}
},
"emails" : {
"address" : "abc123#gmu.edu",
"verified" : true
},
"profile" : {
"companyName" : "comp1",
"flagged" : true,
"phoneNum" : "7778883333"
}}
I want to return and store the value for profile.flagged specifically for the document with _id : XfCZSje7GjynvMZu7. So far I have tried:
db.users.find({_id:'myfi3E4YTf9z6tdgS'},{admin:1})
and
db.users.find({_id: 'myfi3E4YTf9z6tdgS'}, {profile:admin});
I want the query to return true or false depending on the assigned value.
Can someone help? Thanks!
MongoDB queries always return document objects, not single values. So one way to do this is with shell code like:
var flagged =
db.users.findOne({_id: 'myfi3E4YTf9z6tdgS'}, {'profile.flagged': 1}).profile.flagged;
Note the use of findOne instead of find so that you're working with just a single doc instead of the cursor that you get with find.
The correct answer here is the method .distinct() (link here)
Use it like this:
db.users.find({_id:'myfi3E4YTf9z6tdgS'},{admin:1}).distinct('admin')
The result will be: 1 or 0
Related
I'm new to mongo and am trying to do a very simple query in this collection:
{
"_id" : ObjectId("gdrgrdgrdgdr"),
"administrators" : {
"-HGFsfes" : {
"name" : "Jose",
"phone" : NumberLong(124324)
},
"-HGFsfqs" : {
"name" : "Peter",
"phone" : "+43242342"
}
},
"countries" : {
"-dgfgrdg : {
"lang" : "en",
"name" : "Canada"
},
"-grdgrdg" : {
"lang" : "en",
"name" : "USA"
}
}
}
How do I make a query that returns the results of administrators with name like "%Jos%" for example.
What I did until now is this: db.getCollection('coll').find({ "administrators.name": /Jos/});
And variations of this. But every thing I tried returns zero results.
What am I doing wrong?
Thanks in advance!
Your mistake is that administrators is not an array, but an object with fields that are themselves objects with name field. Right query will be
{ "administrators.-HGFsfes.name": /Jos/}
Unfortunatelly this way you're only querying -HGFsfes name field, not other administrator name field.
To achieve what you want, the only thing to do is to replace administrators object by an array, so your document will look like this :
{
"administrators" : [
{
"id" : "-HGFsfes",
"name" : "Jose",
"phone" : 124324
},
{
"id" : "-HGFsfqs",
"name" : "Peter",
"phone" : "+43242342"
}
],
countries : ...
}
This way your query will work.
BUT it will return documents where at least one entry in administrators array has the matching name field. To return only administrator matching element, and not whole document, check this question and my answer for unwind/match/group aggregation pipeline.
You need to use query like this:
db.collection_name.find({})
So if your collection name is coll, then it would be:
db.coll.find({"administrators.-HGFsfes.name": /Jos/});
Look this for like query in mongo.
Also, try with regex pattern like this:
db.coll.find({"administrators..-HGFsfes.name": {"$regex":"Jos", "$options":"i"}}});
It will give you only one result because your data is not an array as below in screenshot:
If you want multiple results, then you need to restructure your data.
Ok, think i've found a better solution for you, with aggregation framework.
Run the following query on your current collection, will return you all administrators with name "LIKE" jos (case insensitive with i option) :
db.test1.aggregate(
[
{
$project: {
administrators:{ $objectToArray: "$administrators"}
}
},
{
$unwind: {
path : "$administrators"
}
},
{
$replaceRoot: {
newRoot:"$administrators"
}
},
{
$match: {
"v.name":/jos/i
}
},
]
);
Output
{
"k" : "-HGFsfes",
"v" : {
"name" : "Jose",
"phone" : NumberLong(124324)
}
}
"k" and "v" are coming from "$objectToArray" operator, you can add a $project stage to rename them (or discard if k value doesn't matter)
Not sure for Robomongo testing but in Studio 3T, formerly Robomongo, you can either copy/paste this query in Intellishell console, or copy/import in aggregation tab, (small icon 'paste from the clipboard').
Hope it helps.
I have a set of mongodb documents with the following structure:
{
"_id" : NUUID("58fbb893-dfe9-4f08-a761-5629d889647d"),
"Identifiers" : {
"IdentificationLevel" : 2,
"Identifier" : "extranet\\test#test.com"
},
"Personal" : {
"FirstName" : "Test",
"Surname" : "Test"
},
"Tags" : {
"Entries" : {
"ContactLists" : {
"Values" : {
"0" : {
"Value" : "{292D8695-4936-4865-A413-800960626E6D}",
"DateTime" : ISODate("2015-04-30T09:14:45.549Z")
}
}
}
}
}
}
How can I make a query with the mongo shell which finds all documents with a specific "Value" (e.g.{292D8695-4936-4865-A413-800960626E6D} in the Tag.Entries.ContactLists.Values path?
The structure is unfortunately locked by Sitecore, so it is not an options to use another structure.
As your sample collection structure show Values is object, it contains only one Value. Also you must check for Value as it contains extra paranthesis. If you want to get Value from given structure try following query :
db.collection.find({
"Tags.Entries.ContactLists.Values.0.Value": "{292D8695-4936-4865-A413-800960626E6D}"
})
How to get all data with condition discountPrice not nil?
if i have object like this inside my mongo db :
when i'm execute this query "db.products.find()", i get result like below :
{
"_id" : ObjectId("123456778"),
"items" : [
{
"discountPrice" : 159200,
"_id" : ObjectId("54697e689857572459444162"),
}
]
},
{
"_id" : ObjectId("847446468"),
"items" : [
{
"discountPrice" : nil,
"_id" : ObjectId("54697e689857572459444162"),
}
]
}
how to get result like this :
{
"_id" : ObjectId("123456778"),
"items" : [
{
"discountPrice" : 159200,
"_id" : ObjectId("54697e689857572459444162"),
}
]
}
how do that?
You can use the $ne operator to match values that are "not equal" to null:
db.products.find( { "items.discountprice": { "$ne": null } } )
I've changed your reference of nil to null because nil is not a valid BSON type. I'm assuming that you are using ruby which uses nil in favor of null. Using your mongodb drivers, I think you'll still be able to actually supply the value nil, but in this language agnostic post the correct value would be null.
You can do it like this:
db.products.find({'items.discountPrice':{ '$ne': null }});
more info on queries can be found here : http://docs.mongodb.org/manual/reference/operator/query/exists/
Edit: updated example with using null.
My MongoDB collection is made up of 2 main collections :
1) Maps
{
"_id" : ObjectId("542489232436657966204394"),
"fileName" : "importFile1.json",
"territories" : [
{
"$ref" : "territories",
"$id" : ObjectId("5424892224366579662042e9")
},
{
"$ref" : "territories",
"$id" : ObjectId("5424892224366579662042ea")
}
]
},
{
"_id" : ObjectId("542489262436657966204398"),
"fileName" : "importFile2.json",
"territories" : [
{
"$ref" : "territories",
"$id" : ObjectId("542489232436657966204395")
}
],
"uploadDate" : ISODate("2012-08-22T09:06:40.000Z")
}
2) Territories, which are referenced in "Map" objects :
{
"_id" : ObjectId("5424892224366579662042e9"),
"name" : "Afghanistan",
"area" : 653958
},
{
"_id" : ObjectId("5424892224366579662042ea"),
"name" : "Angola",
"area" : 1252651
},
{
"_id" : ObjectId("542489232436657966204395"),
"name" : "Unknown",
"area" : 0
}
My objective is to list every map with their cumulative area and number of territories. I am trying the following query :
db.maps.aggregate(
{'$unwind':'$territories'},
{'$group':{
'_id':'$fileName',
'numberOf': {'$sum': '$territories.name'},
'locatedArea':{'$sum':'$territories.area'}
}
})
However the results show 0 for each of these values :
{
"result" : [
{
"_id" : "importFile2.json",
"numberOf" : 0,
"locatedArea" : 0
},
{
"_id" : "importFile1.json",
"numberOf" : 0,
"locatedArea" : 0
}
],
"ok" : 1
}
I probably did something wrong when trying to access to the member variables of Territory (name and area), but I couldn't find an example of such a case in the Mongo doc. area is stored as an integer, and name as a string.
I probably did something wrong when trying to access to the member variables of Territory (name and area), but I couldn't find an example
of such a case in the Mongo doc. area is stored as an integer, and
name as a string.
Yes indeed, the field "territories" has an array of database references and not the actual documents. DBRefs are objects that contain information with which we can locate the actual documents.
In the above example, you can clearly see this, fire the below mongo query:
db.maps.find({"_id":ObjectId("542489232436657966204394")}).forEach(function(do
c){print(doc.territories[0]);})
it will print the DBRef object rather than the document itself:
o/p: DBRef("territories", ObjectId("5424892224366579662042e9"))
so, '$sum': '$territories.name','$sum': '$territories.area' would show you '0' since there are no fields such as name or area.
So you need to resolve this reference to a document before doing something like $territories.name
To achieve what you want, you can make use of the map() function, since aggregation nor Map-reduce support sub queries, and you already have a self-contained map document, with references to its territories.
Steps to achieve:
a) get each map
b) resolve the `DBRef`.
c) calculate the total area, and the number of territories.
d) make and return the desired structure.
Mongo shell script:
db.maps.find().map(function(doc) {
var territory_refs = doc.territories.map(function(terr_ref) {
refName = terr_ref.$ref;
return terr_ref.$id;
});
var areaSum = 0;
db.refName.find({
"_id" : {
$in : territory_refs
}
}).forEach(function(i) {
areaSum += i.area;
});
return {
"id" : doc.fileName,
"noOfTerritories" : territory_refs.length,
"areaSum" : areaSum
};
})
o/p:
[
{
"id" : "importFile1.json",
"noOfTerritories" : 2,
"areaSum" : 1906609
},
{
"id" : "importFile2.json",
"noOfTerritories" : 1,
"areaSum" : 0
}
]
Map-Reduce functions should not be and cannot be used to resolve DBRefs in the server side.
See what the documentation has to say:
The map function should not access the database for any reason.
The map function should be pure, or have no impact outside of the
function (i.e. side effects.)
The reduce function should not access the database, even to perform
read operations. The reduce function should not affect the outside
system.
Moreover, a reduce function even if used(which can never work anyway) will never be called for your problem, since a group w.r.t "fileName" or "ObjectId" would always have only one document, in your dataset.
MongoDB will not call the reduce function for a key that has only a
single value
I have the following items in my collection:
> db.test.find().pretty()
{ "_id" : ObjectId("532c471a90bc7707609a3d4f"), "name" : "Alice" }
{
"_id" : ObjectId("532c472490bc7707609a3d50"),
"name" : "Bob",
"partner_type1" : {
"status" : "rejected"
}
}
{
"_id" : ObjectId("532c473e90bc7707609a3d51"),
"name" : "Carol",
"partner_type2" : {
"status" : "accepted"
}
}
{
"_id" : ObjectId("532c475790bc7707609a3d52"),
"name" : "Dave",
"partner_type1" : {
"status" : "pending"
}
}
There are two partner types: partner_type1 and partner_type2. A user cannot be accepted partner in the both of types. But he can be a rejected partner in partner_type1 but accepted in the another, for example.
How can I build Mongo query that fetches the users that can become partners?
When your user can only be accepted in one partner-type, you should turn it around: Have a field accepted_as:"partner_type1" or accepted_as:"partner_type2". For people who aren't accepted yet, either have no such field or set it to null.
In both cases, your query to get any non-accepted will then be:
{
data.accepted_as: null
}
(null matches both non-existing fields as well as fields explicitly set to null)
For me the logical schema would be this:
"partner : {
"type": 1,
"status" : "rejected"
}
At least that keeps the paths consistent between documents.
So if you want to stay away from using mapReduce type methods to find out "which field" it is on, and otherwise use plain queries and the aggregation pipeline, then don't vary field paths on documents. If you alter the "data" then that is the most consistent form.