I want to avoid to write conditional statements to check parameters before query my database
so that i don't have to write multiple times the similar query but i have to write one single query.
So my first query is
db.results.find( { travel_class : "first" } })
and returns several documents
my second query is
var travel_class_filter = "all";
db.results.find( {'$and' : [ { travel_class_filter : "all"} , { travel_class : "first" } ]})
and returns 0 document instead of the same number of documents of the second query
Do you know why?
You can't test parameters as part of your query, but you can do this by building up your query object programmatically:
var query = {};
if (travel_class_filter != 'all') {
query.travel_class = travel_class_filter;
}
db.results.find(query);
Related
I have 2 different collections and I am trying to query the first collection and take the output of that as an input to the second collection.
var mycursor = db.item.find({"itemId":NumberLong(123)},{"_id":0})
var outp = "";
while(mycursor.hasNext()){
var rec = mycursor.next()
outp = outp + rec.eventId;
}
This query works fine and returns me a list of eventIds.
I have another collection named users, which has eventId field in it. A eventId can repeat in multiple users. So for each eventId I get in the above query I want to get the list of users too.
My query for the second collection would be something like this :
db.users.find({"eventId":ObjectdId("each eventId from above query")},{"_id":0})
My final result would be a unique list of users.
Whell this should basically work ( to a point that is ):
db.events.find({
"eventId": { "$in": db.item.find({
"itemId":NumberLong(123)
}).map(function(doc) { return doc.eventId }) }
})
Or even a bit better:
db.events.find({
"eventId": { "$in": db.item.distinct("eventId",{
"itemId":NumberLong(123) }) }
})
The reason is that "inner query" is evaluated before the outer query is sent. So you get an array of arguments for use with $in.
That is basically the same as doing the following, which translates better outside of the shell:
var events = db.item.distinct("eventId",{ "itemId":NumberLong(123) });
db.events.find({ "eventId": { "$in": events } })
If the results are "too large" however, then your best approach is to loop the initial results as you have done already and build up an array of arguments. Once at a certain size then do the same $in query several times in "pages" to get the results.
But looking for "ditinct" eventId via .distinct() or .aggregate() will help.
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'm using Mongoose.js to perform an $in query, like so:
userModel.find({
'twitter_username': {
$in: friends
}
})
friends is just an array of strings. However, I'm having some case issues, and wondering if I can use Mongo's $regex functionality to make this $in query case-insensitive?
From the docs:
To include a regular expression in an $in query expression, you can
only use JavaScript regular expression objects (i.e. /pattern/ ). For
example:
{ name: { $in: [ /^acme/i, /^ack/ ] } }
One way is to create regular Expression for each Match and form the friends array.
var friends = [/^name1$/i,/^name2$/i];
or,
var friends = [/^(name1|name2)$/i]
userModel.find({"twitter_username":{$in:friends }})
Its a little tricky to do something like that
at first you should convert friends to new regex array list with:
var insesitiveFriends = [];
friends.forEach(function(item)
{
var re = new RegExp(item, "i");
insesitiveFriends.push(re);
})
then run the query
db.test.find(
{
'twitter_username':
{
$in: insesitiveFriends
}
})
I have the sample documents in test collection
/* 0 */
{
"_id" : ObjectId("5485e2111bb8a63952bc933d"),
"twitter_username" : "David"
}
/* 1 */
{
"_id" : ObjectId("5485e2111bb8a63952bc933e"),
"twitter_username" : "david"
}
and with var friends = ['DAvid','bob']; I got both documents
Sadly, mongodb tends to be pretty case-sensitive at the core. You have a couple of options:
1) Create a separate field that is a lowercase'd version of the twitter_username, and index that one instead. Your object would have twitter_username and twitter_username_lc. The non-lowerecase one you can use for display etc, but the lowercase one you index and use in your where clause etc.
This is the route I chose to go for my application.
2) Create a really ugly regex from your string of usernames in a loop prior to your find, then pass it in:
db.users.find({handle:{$regex: /^benhowdle89|^will shaver|^superman/i } })
Note that using the 'starts with' ^ carrot performs better if the field is indexed.
This is my query.
db.final.find( { 'skills.skillList.skill.name' : { $exists : true } } )
I want to write results of this query to a separate collection.
Is it possible?
Fetch results to the client and insert into another collection. It's not possible in the database, without retrieving results.
I am using the $in clause within a mongodb query. However, I want my results to be ordered exactly as my input list used in the $in clause.
The results I get back are in the order the objects were inserted.
All the documentation I have read so far tells me that the sort cursor function only works with document fields. Does this mean that I cannot do the above?
MongoDB does not insert in any particular order. You can sort your results though.
See:
http://www.mongodb.org/display/DOCS/Advanced+Queries#AdvancedQueries-%24in
db.updates.ensureIndex( { ts : 1 } ); // ts == timestamp
var myFriends = myUserObject.friends; // let's assume this gives us an array of DBRef's of my friends
var latestUpdatesForMe = db.updates.find( { user : { $in : myFriends } } ).sort( { ts : -1 } ).limit(10);