I am new to MongoDB. I have a collection called person. I'm trying to get all the records without an _id field with this query:
db.person.find({}{_id:0})
but the error is
syntax error: unexpected {
but if i write
db.person.find()
it works perfectly.
Consider following documents inserted in person collection as
db.person.insert({"name":"abc"})
db.person.insert({"name":"xyz"}
If you want to find exact matching then use query as
db.person.find({"name":"abc"})
this return only matched name documents
If you want all names without _id then use projeciton id query as
db.person.find({},{"_id":0})
which return
{ "name" : "abc" }
{ "name" : "xyz" }
According to Mongodb manual you have little wrong syntax, you forgot to give comma after {}
Try this :
db.person.find({}, { _id: 0 } )
Related
I have to run mongo updates for employee id linked to a supervisorId (That's an oversimplification, but its the gist of what I need). The find uses a list of employee ids, and if the employee id is not found, I have an upsert to add the data. However the issue is I'm not getting the employee id in the upsert. Is it possible to upsert values from the $in array?
Eg:
db.collection.update({
empId:{
$in:['EMP1','EMP2','EMP3']
}
},
{$set: {SupervisorId:'SUP25'}},
{upsert:true},
{multi:true})
If I run this query I get one new doc with _id and SupervisorId:
{
"_id" : ObjectId("5e56ece8b0ae3537d0261ghg"),
"SupervisorId" : "SUP25"
}
I would like to get this in my upserts:
{
"_id" : ObjectId("5e56ece8b0ae3537d0261ghg"),
"SupervisorId" : "SUP25"
"EmpId": "EMP1"
}
I'm still getting comfortable with mongo, but is this even possible with an upsert? If not possible, are there any options for something similar?
MongoDB version 2.9.1
I have a collection where the _id is of the form [message_code]-[language_code] and another where the _id is just [message_code]. What I'd like to do is find all documents from the first collection where the message_code portion of the _id does not appear in the second collection.
Example:
> db.colA.find({})
{ "_id" : "TRM1-EN" }
{ "_id" : "TRM1-ES" }
{ "_id" : "TRM2-EN" }
{ "_id" : "TRM2-ES" }
> db.colB.find({})
{ "_id" : "TRM1" }
I want a query that will return TRM2-EN and TRM-ES from colA. Of course in my live data, there are thousands of records in each collection.
According to this question which is trying to do something similar, we have to save the results from a query against colB and use it in an $in condition in a query against colA. In my case, I need to strip the -[language_code] portion before doing this comparison, but I can't find a way to do so.
If all else fails, I'll just create a new field in colA that contains only the message code, but is there a better way do it?
Edit:
Based on Michael's answer, I was able to come up with this solution:
var arr = db.colB.distinct("_id")
var regexs = arr.map(function(elm){
return new RegExp(elm);
})
var result = db.colA.find({_id : {$nin : regexs}}, {_id : true})
Edit:
Upon closer inspection, the above method doesn't work after all. In the end, I just had to add the new field.
Disclaimer: This is a little hack it may not end well.
Get distinct _id using collection.distinct method.
Build a regular expression array using Array.prototype.map()
var arr = db.colB.distinct('_id');
arr.map(function(elm, inx, tab) {
tab[inx] = new RegExp(elm);
});
db.colA.find({ '_id': { '$nin': arr }})
I'd add a new field to colA since you can index it and if you have hundreds of thousands of documents in each collection splitting the strings will be painfully slow.
But if you don't want to do that you could make use of the aggregation framework's $substr operator to extract the [message-code] then do a $match on the result.
I have a couple of collections, for example;
members
id
name
//other fields we don't care about
emails
memberid
//other fields we don't care about
I want to delete the email for a given member. In SQL I could use a nested query, something like;
delete emails
where memberid in (select id from members where name = "evanmcdonnal")
In mongo I'm trying something like this;
db.emails.remove( {"memberid":db.members.find( {"name":"evanmcdonnal"}, {id:1, _id:0} ) )
But it returns no results. So I took the nested query and ran it on it's own. The issue I believe is that it returns;
{
"id":"myMadeUpId"
}
Which - assuming inner queries execute first - gives me a query of;
db.emails.remove( {"memberid":{ "id""myMadeUpId"} )
When really I just want the value of id. I've tried using dictionary and dot notation to access the value of id with no luck. Is there a way to do this that is similar to my attempted query above?
Let's see how you'd roughly translate
delete emails where memberid in (select id from members where name = "evanmcdonnal")
into a set of mongo shell operations. You can use:
db.members.find({ "name" : "evanmcdonnal" }, { "id" : 1 }).forEach(function(doc) {
db.emails.remove({ "memberid" : doc.id });
});
However, this does one remove query for each result document from members. You could push the members result ids into an array and use $in:
var plzDeleteIds = db.members.find({ "name" : "evanmcdonnal" }, { "id" : 1 }).toArray();
db.emails.remove({ "memberid" : { "$in" : plzDeleteIds } });
but that could be a problem if plzDeleteIds gets very, very large. You could batch. In all cases we need to do multiple requests to the database because we are querying multiple collections, which always requires multiple operations in MongoDB (as of 2.6, anyway).
The more idiomatic way to do this type of thing in MongoDB is to store the member information you need in the email collection on the email documents, possibly as a subdocument. I couldn't say exactly if and how you should do this since you've given only a bit of your data model that has, apparently, been idealized.
As forEach() way didn't work for me i solved this using:
var plzDeleteIds = db.members.find({ "name" : "evanmcdonnal" }, { "id" : 1 }).toArray();
var aux = plzDeleteIds["0"];
var aux2 = aux.map(function(u) { return u.name; } );
db.emails.remove({ "memberid" : { "$in" : aux2 } });
i hope it help!
I do not believe what you are asking for is possible. MongoDB queries talk to just one collection -- there is no syntax to go cross-collection.
However, what about the following:
The name in members does not seem to be unique. If you were to delete emails from the "emails' collection using name as the search attribute, you might have a problem. Why not store the actual email address in the email collection? And store email address again in the members collection. When your user logs in, you will have retrieved his member record -- including the email address. When you want to delete his emails, you already have his email and you can do:
db.emails.remove({emailAddress: theActualAddress))
Does that work?
Is there any way to remove all the documents except one from a collection based on condition.
I am using MongoDB version 2.4.9
You can do this in below way,
db.inventory.remove( { type : "food" } )
Above query will remove documents with type equals to "food"
To remove document that not matches condition you can do,
db.inventory.remove( { type : { $ne: "food" } } )
or
db.inventory.remove( { type : { $nin: ["Apple", "Mango"] } } )
Check here for more info.
To remove all documents except one, we can use the query operator $nin (not in) over a specified array containing the values related to the documents that we want to keep.
db.collections.remove({"field_name":{$nin:["valueX"]}})
The advantage of $nin array is that we can use it to delete all documents except one or two or even many other documents.
To delete all documents except two:
db.collections.remove({"field_name":{$nin:["valueX", "valueY"]}})
To delete all documents except three:
db.collections.remove({"field_name":{$nin:["valueX", "valueY", "valueZ"]}})
Query
db.collection.remove({ "fieldName" : { $ne : "value"}})
As stated above by Taha EL BOUFFI, the following worked for me.
db.collection.remove({"fieldName" : { $nin: ["value"]}});
2 days old to Mongo, so bear with me.
I have a collection from which, I only want to retrieve specific values contingent to another key existing in the MongoDB environment.
Here is what I am doing:
db.results.find({'someKeyThatShouldExist':{$exists:true}}, {"parentKey.childKey.theKeyWoseValueIwant":1}
This yields data in the following format for me:
{ "_id" : ObjectId("532a2c2b6803fa486b8b456a"), "parentKey" : { "childKey" : { "theKeyWhoseValueIWant" : 102982577 }}}.....
Now, all I really want is the value 102982577, not everything else.
How can I do this ?
You can suppress the _id by adding _id:0 to the projection criteria.
db.results.find(
{"someKeyThatShouldExist":{$exists:true}},
{_id:0, "parentKey.childKey.theKeyWoseValueIwant":1}
)
To get just the value, you could do something like:
db.results.find(
{"someKeyThatShouldExist":{$exists:true}},
{_id:0, "parentKey.childKey.theKeyWoseValueIwant":1}
)[0].parentKey.childKey.theKeyWoseValueIwant