So i need some advice as to what i'm doing incorrectly.
My database is setup up exactly like a file system consisting of folders and files.
It begins with a folder, but can have a relatively infinite number of subfolders and or files.
{
"name":"folder1",
"uniqueID":"zzz0",
"subcontents": [ {"name":"subfolder1", "uniqueID":"zzz1"},
{"name":"subfile1", "uniqueID":"zzz2"},
{"name":"subfile2", "uniqueID":"zzz3"},
{"name":"subfolder2", "subcontents": [...etc...], "uniqueID":"zzz4"},
]
}
Each folder/file document have a uniqueID so that I can reference to it (seen above zzz#). My question is, can I make a mongoDB query to pull out only a single document?
Like say for example db.fileSystemCollection.find({"uniqueID":"zzz4"}) and it would give me the following result? Do i have to use indexes to do this? I've been trying but the query returns empty every time.
intended result ---> {"name":"subfolder2", "subcontents": [...etc...], "uniqueID":"zzz4"}
[EDIT]
Based on the responses below, I will consider an XML database instead on mongoDB. The json structure cant be rearranged to work with MongoDB (too much data).
Short answer is no, as it's stated by Chris.
Your embedded representation of a tree is really good for intuitive understanding (and implementation as well). But if you want to allow effective searches on your tree using indices in MongoDB, you might consider another ways for tree storage. A bunch of ways is listed at http://docs.mongodb.org/manual/tutorial/model-tree-structures/
Please keep in mind that every representation has its own pros and cons depending on your access patterns.
Since for filesystem-like structure it's likely to have the ability to find all the sub contents of a given folder, you may use child references pattern for this:
{
"name":"folder1",
"uniqueID":"zzz0",
"subcontents": [ "zzz1",
"zzz2",
"zzz3",
"zzz4"
]
}
{
"name":"subfolder1",
"uniqueID":"zzz1"
}
...
No; searching for {uniqueID: "zzz4"} will only get you documents whose top-level uniqueID matches.
What you probably want is to maintain an array on the document which lists all the unique IDs in that tree. So your document would be:
{
"name":"folder1",
"uniqueID":"zzz0",
"idList": ["zzz0", "zzz1", "zzz2", "zzz3", "zzz4"],
"subcontents": [ {"name":"subfolder1", "uniqueID":"zzz1"},
{"name":"subfile1", "uniqueID":"zzz2"},
{"name":"subfile2", "uniqueID":"zzz3"},
{"name":"subfolder2", "subcontents": [...etc...], "uniqueID":"zzz4"},
]
}
Then you can index that:
db.fileSystemCollection.ensureIndex({"idList": 1})
Then you can find on it:
db.fileSystemCollection.find({"idList": "zzz4})
That'll return you those documents.
As an aside, if you're trying to store files in Mongo, have you looked at GridFS?
Related
How to find partial search?
Now Im trying to find
db.content.find({$text: {$search: "Customer london"}})
It finds all records matching customer, and all records matching london.
If I am searching for a part of a word for example lond or custom
db.content.find({$text: {$search: "lond"}})
It returns an empty result. How can I modify the query to get the same result like when I am searching for london?
You can use regex to get around with it (https://docs.mongodb.com/manual/reference/operator/query/regex/). However, it will work for following :
if you have word Cooking, following queries may give you result
cooking(exact matching)
coo(part of the word)
cooked(The word containing the english root of the document word, where cook is the root word from which cooking or cooked are derived)
If you would like to go one step further and get a result document containing cooking when you type vooking (missplled V instead of C), go for elasticsearch.
Elasticsearch is easy to setup, has extremely powerful edge-ngram analyzer which converts each words into smaller weightage words. Hence when you misspell, you will still get a document based on score elasticsearch gives to that document.
You can read about it here : https://www.elastic.co/guide/en/elasticsearch/reference/current/analysis-edgengram-tokenizer.html
it will always return the empty array for partial words like when you are searching for lond to get this type of text london..
Because it take full words and search same as that they are ..
Not achive same results like :-
LO LON LOND LONDO LONDON
Here you may get help from ELASTIC-SEARCH . It is quite good for full text search when implement with mongoDB.
Refrence : ElasticSearch
Thanks
The find all is to an Array
clientDB.collection('details').find({}).toArray().then((docs) =>
I now used the str.StartWith in a for loop to pick out my record.
if (docs[i].name.startsWith('U', 0)) {
return console.log(docs[i].name);
} else {
console.log('Record not found!!!')
};
This may not be efficient, but it works for now
Lets say I have a document structured like that :
datas: {
foo: {
...
keytoupdate: [...]
},
whatever: {
...
keytoupdate: [...]
},
anystring: {
...
keytoupdate: [...]
},
...: {
...
keytoupdate: [...]
}
}
I know that :
Each direct child property of the "datas" document has a "keytoupdate" field.
The direct child properties of the "datas" document varies from case to case: not necessarily the same name, neither the same number.
I want to update each "keytoupdate" fields, no matter how many of them there are.
The question is: How can I do that ? Is there any magic operator like $ that does the same job for Array ?
Thank you !
I'll answer my own question : there is no way to do that, we can't play with dynamic keys, just forget about it ! But there are 2 workarounds :
The best solution, as suggested by #chridam, is to redesign the schema to make an array of objects, where the keys are parts of the arrays, you can see this question for more details.
If you can't, the other (but not good) solution is to make a request for each field that might be in your document, instead of trying to do this in one request. This is a very bad solution, especially if your document may have lots of fields, and you have to known which fields that could be in your documents. This is a bad solution, absolutely not optimized, but it has the merit of being simple to implement
I know this has been asked before, but I have yet to find a solution that works efficiently. I am working with the MongoDB C# driver, though this is more of a general question about MongoDB operations.
I have a document structure that looks something like this:
field1: value1
field2: value2
...
users: [ {...user 1 subdocument...}, {...user 2 subdocument...}, ... ]
Some facts:
Each user subdocument includes further sub-arrays & subdocuments (so they're fairly complex).
The average users array only contains about 5 elements, but in the worst case can surpass 100.
Several thousand update operations on multiple users may be conducted per day in this system, each on one document at a time. Larger arrays will receive more frequent updates due to their data size.
I am trying to figure out how to do this efficiently. From what I've heard, you cannot directly set several array elements to new values all at once, so I had to try something else.
I tried using the $pullAll / $AddToSet + $each operations to remove the old array and replace it with a modified one. I am aware that $pullall can remove only the elements that I need as well, but I would like to preserve the order of elements.
The C# code:
try
{
WriteConcernResult wcr = collection.Update(query,
Update.Combine(Update.PullAll("users"),
Update.AddToSetEach("users", newUsers.ToArray())));
}
catch (WriteConcernException wce)
{
return wce.Message;
}
In this case newUsers is aList<BsonValue>converted to an array. However I am getting the following exception message:
Cannot update 'users' and 'users' at the same time
By the looks of it, I can't have two update statements in use on the same field in the same write operation.
I also tried Update.Set("users", newUsers.ToArray()), but apparently the Set statement doesn't work with arrays, just basic values:
Argument 2: cannot convert from 'MongoDB.Bson.BsonValue[]' to 'MongoDB.Bson.BsonValue'
So then I tried converting that array to a BsonDocument:
Update.Set("users", newUsers.ToArray().ToBsonDocument());
And got this:
An Array value cannot be written to the root level of a BSON document.
I could try replacing the whole document, but that seems like overkill and definitely not very efficient.
So the only thing I can think of now is to run two separate write operations: one to remove the unwanted old users and another to replace them with their newer versions:
WriteConcernResult wcr = collection.Update(query, Update.PullAll("users"));
WriteConcernResult wcr = collection.Update(query, Update.AddToSetEach("users", newUsers.ToArray()));
Is this my best option? Or is there another, better way of doing this?
Your code should work with a minor change:
Update.Set("users", new BsonArray(newUsers));
BsonArray is a BsonValue, where as an array of documents is not and we don't implicitly convert arrays like we do other primitive values.
this extension method solve my problem:
public static class MongoExtension
{
public static BsonArray ToBsonArray(this IEnumerable list)
{
var array = new BsonArray();
foreach (var item in list)
array.Add((BsonValue) item);
return array;
}
}
My application has the following stack:
Sinatra on Ruby -> MongoMapper -> MongoDB
The application puts several entries in the database. In order to crosslink to other pages, I've added some sort of syntax. e.g.:
Coffee is a black, caffeinated liquid made from beans. {Tea} is made from leaves. Both drinks are sometimes enjoyed with {milk}
In this example {Tea} will link to another DB entry about tea.
I'm trying to query my mongoDB about all 'linked terms'. Usually in ruby I would do something like this: /{([a-zA-Z0-9])+}/ where the () will return a matched string. In mongo however I get the whole record.
How can I get mongo to return me only the matched parts of the record I'm looking for. So for the example above it would return:
["Tea", "milk"]
I'm trying to avoid pulling the entire record into Ruby and processing them there
I don't know if I understand.
db.yourColl.aggregate([
{
$match:{"yourKey":{$regex:'[a-zA-Z0-9]', "$options" : "i"}}
},
{
$group:{
_id:null,
tot:{$push:"$yourKey"}
}
}])
If you don't want to have duplicate in totuse $addToSet
The way I solved this problem is using the string aggregation commands to extract the StartingIndexCP, ending indexCP and substrCP commands to extract the string I wanted. Since you could have multiple of these {} you need to have a projection to identify these CP indices in one shot and have another projection to extract the words you need. Hope this helps.
here's my model:
class Person
acts_as_tree
end
i relate multiple objects as a tree:
P1
|
---
| |
P1.1 P1.2
|
---
| |
P1.1.1 P1.1.2
here if i need to retreive P1.1.1 i need to write a query that effectively asks:
get me the Person with name P1.1.1 and path (given by acts_as_tree) [P1, P1.1].
querying by just name is not enough as i can have similar named people at multiple paths.
how do i do this?
> db.people.find({name: 'P1.1.1'})
above snippet will show me the path attribute correctly as expected, but i cannot query by that path.
> db.people.find({name: 'P1.1.1', path: [{name: 'P1'}, {name: 'P1.1'}]})
doesn't work. neither does:
> db.people.find({name: 'P1.1.1', path: [db.people.find({name: 'P1'}),
db.people.find({name: 'P1.1'})]})
but that explains what i'm trying to do.
One of the ways where you can query something like as follows:
db.people.find({name : 'Joe', 'path' : { $all : [ObjectId("4e0fcf1722b7a9439200002e"), ObjectId("4e0fcf1622b7a9439200002b")]}})
However the drawback I think of this is:
You don't get to substitute the object relations/joins directly in mongo shell. You have to use the ObjectId object
The $all clause does not necessitate that the order of path is strictly same, which means that a person with name "Joe" referenced by path "hometown/town/" would come up as well as "Joe" from "tome/hometown".
I would presume that second one may be a deal breaker. Also, I am presuming that mongoid in several cases pass across the query options directly to mongodb (or atleast there are ways to do that). Hence it should be possible to do a search in ruby code using given query above.
Nevertheless, I'll do some more re-search on this and post my findings back.
Hope it helps.
Edit
To alleviate the second problem above there is also another way to query a person with a specific path. Find it below:
db.people.find({name : 'Joe', 'path.0' :ObjectId("4e0fcf1722b7a9439200002e"), 'path.1':ObjectId("4e0fcf1622b7a9439200002b")})
This would ensure that path is exactly what you are looking for. However this works in mongodb shell and you may still need to figure out, how mongoid can run an equivalent of this. Plus you may have to construct this query dynamically to create a path for person and that (for deep nested people) may just become long and ugly.
I would suggest skimming through following links on mongodb documentation to get a better understanding.
http://www.mongodb.org/display/DOCS/Dot+Notation+%28Reaching+into+Objects%29
http://www.mongodb.org/display/DOCS/Advanced+Queries#AdvancedQueries-ValueinanArray
I hope this is what you were looking for.