I have a simple collection like below
> db.test.save({first:"Ab"})
> db.test.find()
{ "_id" : ObjectId("518a1524f635dc8bb092e1ac"), "first" : "Ab" }
I want to add a new field called 'fl' which holds the first letter of the field "first".
I tried this
> db.test.update({},{"$set":{"fl":{"$substr":["$first",0,1]}}})
not okForStorage
But I get the exception "not okForStorage" as you can see.
Any suggestions, workarounds?
Possibly a duplicate of Multiply field by value in Mongodb but here's a workaround:
db.test.find().forEach(function(e) {
e.fi = e.first[0];
db.save(e);
});
Related
I am new to querying dbs and especially mongodb.If I run :
db.<customers>.find({"contact_name: Anny Hatte"})
I get:
{
"_id" : ObjectId("55f7076079cebe83d0b3cffd"),
"company_name" : "Gap",
"contact_name" : "Anny Hatte",
"email" : "ahatte#gmail.com"
}
I wish to get the value of the "_id" attribute from this query result. How do I achieve that?
Similarly, if I have another collection, named items, with the following data:
{
"_id" : ObjectId("55f7076079cebe83d0b3d009"),
"_customer" : ObjectId("55f7076079cebe83d0b3cfda"),
"school" : "St. Patrick's"
}
Here, the "_customer" field is the "_id" of the customer collection (the previous collection). I wish to get the "_id", the "_customer" and the "school" field values for the record where "_customer" of items-collection equals "_id" of customers-collection.
How do I go about this?
I wish to get the value of the "_id" attribute from this query result.
How do I achieve that?
The find() method returns a cursor to the results, which you can iterate and retrieve the documents in the result set. You can do this using forEach().
var cursor = db.customers.find({"contact_name: Anny Hatte"});
cursor.forEach(function(customer){
//access all the attributes of the document here
var id = customer._id;
})
You could make use of the aggregation pipeline's $lookup stage that has been introduced as part of 3.2, to look up and fetch the matching rows in some other related collection.
db.customers.aggregate([
{$match:{"contact_name":"Anny Hatte"}},
{$lookup:{
"from":"items",
"localField":"_id",
"foreignField":"_customer",
"as":"items"
}}
])
In case you are using a previous version of mongodb where the stage is not supported, then, you would need to fire an extra query to lookup the items collection, for each customer.
db.customers.find(
{"contact_name":"Anny Hatte"}).map(function(customer){
customer["items"] = [];
db.items.find({"_customer":customer._id}).forEach(function(item){
customer.items.push(item);
})
return customer;
})
My question was, how to find all title of the releases contains the artist name
Here is the 2 Documents for just an example:
`{"release" : {"title" : "DEF day",
"artists" : {"artist" : {"role" : "1","name" : "DEF"}}}
}
{ "release" : {"title" : "XYZ day",
"artists" : {"artist" : {"role" : "1","name" : "KYC"}}}
}`
when i run this following query:
`db.test.find({$where:
"this.release.title.indexOf(this.release.artists.artist.name) > -1"
})`
I get the result "DEF day", so it works excellent!!
BUT, once i run this query to my original data(same format like above) i get this:
`error: {
"$err" : "TypeError: Object 30 has no method 'indexOf'\n at _funcs1 (_funcs1:1:49) near 'his.release.artists.a' ",
"code" : 16722
}`
My collection is big (8GB+).
Just looking at the above error, can anyone tell what may be the problem with the above query? your answer is appreciated. please.
The problem is in the data itself. Somewhere you have a document, where the title field is not String, thus invoking indexOf on the whole data, on that particular document you get the error.
One thing you can do to solve this problem is first of all find that erroneous document. To do that you will need to run a simple query.
MongoDB has the $type operator, which matches fields by their type. And here is the list of all available BSON types a MongoDB document field can have.
The id of String BSON type is 2, thus you need all the documents that have type of title field not equal to 2.
find({field: {$not: {$type: <BSON type>}}}))
After finding the document(s), you may fix or remove them.
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 $match on the result.
It seems that i can go further than one subdocument if i want to add it dynamicaly, here is the code:
db.users.update({"pup.cmn.id":id}, {"$addToSet":{"pup.cmn":{"abus":email}}})
this give error:
OperationFailure: can't append to array using string field name: cmn
then, if i add positional element i get this:
db.users.update({"pup.cmn.id":id}, {"$addToSet":{"pup.$.cmn":{"abus":email}}})
"cmn" :
[
{
"fto" : ObjectId("5190e8a53a5f3a0c102af045")
"id" : "14.05.2013 12:29:53"
},
{
"abus" : "u...#example.com"
}
]
so as you can see, it will add it in the same level, and i dont want that, because the application will get errors.
It seems that Mongodb for the time of writing (2.4.x) have not this feature, there is a ticket:
https://jira.mongodb.org/browse/SERVER-831
How to get position (index) of selected document in mongo collection?
E.g.
this document: db.myCollection.find({"id":12345})
has index 3 in myCollection
myCollection:
id: 12340, name: 'G'
id: 12343, name: 'V'
id: 12345, name: 'A'
id: 12348, name: 'N'
If your requirement is to find the position of the document irrespective of any order, that is not
possible as MongoDb does not store the documents in specific order.
However,if you want to know the index based on some field, say _id , you can use this method.
If you are strictly following auto increments in your _id field. You can count all the documents
that have value less than that _id, say n , then n + 1 would be index of the document based on _id.
n = db.myCollection.find({"id": { "$lt" : 12345}}).count() ;
This would also be valid if documents are deleted from the collection.
As far as I know, there is no single command to do this, and this is impossible in general case (see Derick's answer). However, using count() for a query done on an ordered id value field seems to work. Warning: this assumes that there is a reliably ordered field, which is difficult to achieve in a concurrent writer case. In this example _id is used, however this will only work with a single writer case.:
MongoDB shell version: 2.0.1
connecting to: test
> use so_test
switched to db so_test
> db.example.insert({name: 'A'})
> db.example.insert({name: 'B'})
> db.example.insert({name: 'C'})
> db.example.insert({name: 'D'})
> db.example.insert({name: 'E'})
> db.example.insert({name: 'F'})
> db.example.find()
{ "_id" : ObjectId("4fc5f040fb359c680edf1a7b"), "name" : "A" }
{ "_id" : ObjectId("4fc5f046fb359c680edf1a7c"), "name" : "B" }
{ "_id" : ObjectId("4fc5f04afb359c680edf1a7d"), "name" : "C" }
{ "_id" : ObjectId("4fc5f04dfb359c680edf1a7e"), "name" : "D" }
{ "_id" : ObjectId("4fc5f050fb359c680edf1a7f"), "name" : "E" }
{ "_id" : ObjectId("4fc5f053fb359c680edf1a80"), "name" : "F" }
> db.example.find({_id: ObjectId("4fc5f050fb359c680edf1a7f")})
{ "_id" : ObjectId("4fc5f050fb359c680edf1a7f"), "name" : "E" }
> db.example.find({_id: {$lte: ObjectId("4fc5f050fb359c680edf1a7f")}}).count()
5
>
This should also be fairly fast if the queried field is indexed. The example is in mongo shell, but count() should be available in all driver libs as well.
This might be very slow but straightforward method. Here you can pass as usual query. Just I am looping all the documents and checking if condition to match the record. Here I am checking with _id field. You can use any other single field or multiple fields to check it.
var docIndex = 0;
db.url_list.find({},{"_id":1}).forEach(function(doc){
docIndex++;
if("5801ed58a8242ba30e8b46fa"==doc["_id"]){
print('document position is...' + docIndex);
return false;
}
});
There is no way that MongoDB can return this as it does not keep documents in order in the database, just like MySQL f.e. doesn't name row numbers.
The ObjectID trick from jhonkola will only work if only one client creates new elements, as the ObjectIDs are generated on the client side, with the first part being a timestamp. There is no guaranteed order if different clients talk to the same server. Still, I would not rely on this.
I also don't quite understand what you are trying to do though, so perhaps mention that in your question? I can then update the answer.
Restructure your collection to include the position of any entry i.e {'id': 12340, 'name': 'G', 'position': 1} then when searching the database collection(myCollection) using the desired position as a query
The queries I use that return the entire collection all use sort to get a reproducible order, find.sort.forEach works with the script above to get the correct index.