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#mongodb logs for Monday the 13th of October, 2014
[00:05:19] <Synt4x`> so I would do something like db.mydb.aggregate( [ { $group:{ date:{'date':date} }, person{$addToSet:person} }] )? and I do this while iterating through my Person list?
[00:09:12] <Synt4x`> actually I found an alternative I think, .distinct('person',{'date':date}) should work
[00:13:21] <Synt4x`> @joannac yea I did dates = .distinct('date') , for c in dates: people = .distinct('person', {'date':c}) , and then will do a for x in people , .findOne('person':x) and have the data I want
[00:14:09] <Boomtime> that will work, but so will the single operation I quoted for the aggregation pipeline
[00:15:52] <Boomtime> it gives you a cursor, similar to a find, but each document in it looks like this: { _id: [date], person: [ 'name1', 'name2', etc ] }
[00:16:54] <Synt4x`> ah ok awesome, and if I wanted to add 'score' to that as well I would just do: $group: { _id: "$date", person: { $addToSet: "$person" }. {$addToSet: "$score"} }
[00:17:13] <Boomtime> the person array in each document is the distinct list of names that occurred on that date
[00:18:18] <Synt4x`> but if a person has multiple entries under one date, i.e. "Joe Smith", "2014/9/25", "50" , "Joe Smith", "2014/9/25", "50", and so on (same score, same person, multiple entries)
[00:18:20] <Boomtime> this is the first time you've mentioned 'score' so I don't know what it is, but what you've written is not valid syntax
[00:24:23] <Boomtime> you probably want this to include score: $group: { _id: "$date", person: { $addToSet: { name: "$person", score: "$score" } } }
[00:25:36] <Boomtime> if you have the same name on the same date twice but with different scores then it will appear both times in the array for that date
[00:27:21] <Synt4x`> thats actually perfect and exactly what I'm looking for, thanks very much for this!
[00:43:01] <Synt4x`> ok sorry having one more issue, I set dataSet = db.mydb.aggregate() , and then when I do for c in dataSet and print c it shows "ok" , "result"
[00:43:39] <Synt4x`> I was expecting to see {'_id':date, 'person' [{name, score}, {name, score}] , etc.
[00:45:26] <Boomtime> in a driver you will have to do in code what the shell does by guessing what you mean
[00:55:49] <Synt4x`> still here Boomtime? 1 more question now that I have it working properly, what do I do if I want to add non-distinct elements to this,for instance, opponent_score (which will be repeating almost 100% vs all)
[00:58:25] <Boomtime> I can't really tell from your description what you want to achieve, but you would be better off learning the numerous other powerful operators available in the aggregation pipeline
[00:58:27] <Boomtime> http://docs.mongodb.org/manual/reference/operator/aggregation-pipeline/#pipeline-aggregation-stages
[00:59:38] <Boomtime> if you can't figure it out, then you should construct a set of example docs and desired output as a gist on github along with what you've tried
[01:00:30] <Boomtime> you'll get a better response here if you know exactly what you want and can show you've put some effort in already
[01:04:29] <Synt4x`> yea ok I just assumed you would know easily, it's tough switching from my original method to this on the fly and the documentation on $addtoSet doesn't really give many unique examples
[01:06:50] <Boomtime> i would need to understand what you want first, there are many possibilities, put some example documents in a gist along with the output you want
[01:12:18] <Synt4x`> I want it sorted how we already have {Date, (Name, Score) --- the pairing of (name,score) is distinct}, but I also would like to have opponent score listed with each record, and that does not have to be distinct, in fact it will very often be the same repeating with each record, but has the potential to be different
[01:14:24] <Boomtime> i think you need to provide an example output.. i honestly can't tell what you are expecting
[01:15:26] <Boomtime> you need to understand what will happen in all cases, so ensure your sample data covers all possibilities
[01:16:44] <Boomtime> also, those permutations plus whatever complexities the addition of 'opponent score' introduces
[01:24:42] <Synt4x`> ok, so this would be a sample input: http://pastebin.com/FGe5HTFt , this would be an expected output: http://pastebin.com/mKUTXHKr
[01:25:26] <Synt4x`> You can see Robert's score of 83.25 does not repeat (it's inconsequential which of the opponent scores vs the 83.25 it chooses), but he has 2 scores on 2014/09/25 because he has 83.25 and 99.25 so they both are included
[01:29:53] <Synt4x`> basically of all of the ones that get chosen from the original date, (name, score) $group, I just want to add the opponent_score, whatever that was in that instance, to each of those (name, score)
[01:35:55] <Boomtime> you're unique combination is date,name,score and then you simply want whatever happens to be the first opponent-score to be seen on each combination
[01:38:51] <Boomtime> ok, you probably want to do 2 pipeline stages - the first to de-duplicate opponent-scores, i.e group on the combination of date,name,score and select op-score using $first
[01:40:47] <Boomtime> you can then $group again on $_id.date (as it will have become) and now you build your array as before, including op-score knowing it won't cause any new duplicates
[01:45:52] <Synt4x`> will the method you describe make all records on 2014/09/25 for example show the same opponent score? like is it selecting one opp_score ($first) and applying it to all of them
[01:53:41] <Boomtime> it will make all records that have the same value for a given combination of date,name,score have the same op-score
[01:56:52] <Synt4x`> ok, so it should look like dataSet = db.mydb.aggregate({'$group': { '_id': "$date", 'person': { '$addToSet': { 'name': "$opponent", 'score': "$score" }, 'opponent_score':{ '$first':'opponent_score} } }})
[01:57:28] <Synt4x`> trying to read from the $first page on mongodb docs, but again it only gives one example, so I think this should be correct but not 100%
[01:58:57] <Boomtime> i don't actually know what your given pipeline will produce - you should try it and see
[02:00:24] <Boomtime> aggregation is a pipeline, you are using a grand total of 1 stage at the moment, but you can use as many as you like
[02:00:43] <culthero> How do you prevent a mongodb shard from running out of disk space? It has a TTL collection but the volume over a short time increased
[02:00:51] <Synt4x`> ok I think I understand the grouping concept, I guess what I don't fully grasp is where to include the $first to make it non-unique/required
[02:01:24] <Synt4x`> I am assuming if I put it in the original one, it will also be required to be unique, like eahc of those were
[02:02:51] <Boomtime> your second pipeline stage contains something like the 'original' we came up with
[02:03:09] <Synt4x`> ok, I will spend a few hours toying with it now, thanks for all of your help / explanations. Tough for me to learn a new concept from scratch, it seems like the documentation ont his wasn't very thorough
[02:04:25] <Boomtime> culthero: it sounds like you should be using a capped-collection not a TTL index
[02:06:14] <Boomtime> but that isn't replica-sets, nor does it prevent you from using it on a sharded-cluster, just that the collection cannot be sharded
[02:06:41] <Boomtime> capped-collections have always been available on replica-sets, it's kind of essential
[02:07:17] <culthero> it is i/o bound, once it gets to something like 2-3m documents it throttles the disk
[02:17:17] <Boomtime> oh well, if you have size requirements you will probably have to track them yourself
[02:18:21] <culthero> sure but right now for my setup, 3x config servers + 4 shards + mongos instance is pretty cheap
[02:18:26] <Boomtime> the ttl index is a good idea, but it doesn't respond to size - you need a feedback loop to shorten the ttl if the size starts growing
[02:25:02] <culthero> so I will probably run a 3rd node app on my mongos instance that reads something like df -h from each of the shards, and if it drops below like 4gb to reduce the TTL to oldest record - 5 hours
[02:26:18] <Boomtime> be careful with changing TTL significantly - it can cause a sudden mass delete that cause performance impact to other ops
[03:27:03] <Synt4x`> sigh still not able to do this without a TON of iteration, and I know there is a better way now with the aggregation. Hackhands has had me waiting ~1 hour with no expert yet, so if anybody here is great with aggregation and wants to make some $ I will paypal you what I would have paid them ($1/minute)
[05:03:13] <Synt4x`> db.hu_db.aggregate({'$group': { '_id': "$date", 'person': { '$addToSet': { 'name': "$opponent", 'score': "$score", 'opponent_score':'$our_score' } } }})
[05:03:52] <Synt4x`> this gets me what I want except that if (name, score) is the same, but (opponent_score) is different, it will have 2 entries for (name, score), I want it to only have 1
[05:04:05] <Synt4x`> I need (name, score) to be unique, with opponent score just included as extra data (doesnt have to be unique)
[05:04:49] <Synt4x`> it's a bit confusing, "opponent_score" is actually me playing, thats my score. "Score" is the score of my opponent who I played against, but this entry is for their result
[05:08:24] <joannac> you have 2 entries where (name, score) is the same, but (opponent_score) is different
[05:10:17] <Synt4x`> because over time I think it should average out, basically I played 2 "teams" that day, he could have easily been matched vs either one of my teams, so which one it compares him to on the calculates is indifferent
[05:10:31] <Synt4x`> but I still want to do a comparison as far as his %difference from my team, and what percentile he falls in for the day
[05:17:24] <Synt4x`> so I've never done 2 stages, I just do: db.hu_db.aggregate( {'$group': { '_id': "$date", 'person': { '$addToSet': { 'name': "$opponent", 'score': "$score"} } }}, {'$group': { '_id': "$date", 'person': { '$addToSet': { 'name': "$opponent", 'score': "$score", 'opponent_score':'$our_score' } } }} )
[05:21:04] <joannac> You need to use $first http://docs.mongodb.org/manual/reference/operator/aggregation/first/
[05:22:22] <Synt4x`> hrmm sigh this cycle again, I tried 2 hours earlier with Boomtime to do this, and have spent $50 on hackhands/codementor with people who didnt get it right :-p
[05:22:39] <Synt4x`> how would I actually use first in my group statement, I tried earlier and inside, outside, etc.
[05:23:36] <Synt4x`> this was my most recent: dataSet = db.hu_db.aggregate({'$group': { '_id': "$date", 'opponent': { '$addToSet': { 'name': "$opponent", 'score': "$score",'opp_score':{'$first':'$our_score'} } } }})
[05:23:45] <Synt4x`> pymongo.errors.OperationFailure: command SON([('aggregate', u'hu_db'), ('pipeline', [{'$group': {'_id': '$date', 'opponent': {'$addToSet': {'score': '$score', 'name': '$opponent', 'opp_score': {'$first': '$our_score'}}}}}])]) failed: exception: invalid operator '$first'
[05:25:18] <Synt4x`> wowwwww im so dumb, I tried so many times to add it outside of the 'addToSet'{} brackets but inside of the opponent{} ones and it never worked
[05:29:02] <Synt4x`> so if there is no $sort at all before first, it will just grab whatever one the dictionary decides is first? (also unsorted since it's a dict I believe)
[06:07:36] <Synt4x`> eek, @joannac are you still here , I thought it worked perfectly but it's not like I intended it
[06:09:31] <Synt4x`> it only attaches 1 opp_score at the end of the entire list, so it's like {date, {name, score}, {name, score}..., opp_score}, and ideally I want it to have {date, {name, score, opp_score}, {name, score, opp_score}} , I'm just indifferent when there are 1 score (2x) with 2 diff opp_scores, then it can choose either one, doesnt matter
[06:10:57] <Synt4x`> hrmm I guess I misunderstood it then, let me take a look at some two-group stage examples to see if I can see how
[06:13:26] <Synt4x`> this page: http://docs.mongodb.org/manual/reference/operator/aggregation/group/ only has examples of them using 1 group at a time, do I do them both within the same aggregate command?
[06:13:54] <joannac> http://docs.mongodb.org/manual/tutorial/aggregation-zip-code-data-set/#return-average-city-population-by-state
[06:15:13] <Synt4x`> ok yea I read this, so I guess my question is this then, by the looks of the example, if we group on date, (name, score), it gets rid of all of the other data, so if we re-group on that group, we don't have opp_score anymore
[06:16:12] <joannac> see how they're grouping on (city, state) and yet they're also operating on the population
[06:19:35] <Synt4x`> sorry if I'm being obtuse here, but the first $group in both examples are: $group{ {state, city}, population }, they both have all 3 items in their first group command
[06:21:08] <Synt4x`> in the example, Return Largest and Smallest Cities by State , also in Return Average City Population by State
[06:21:19] <Synt4x`> { $group : { _id : { state : "$state", city : "$city" }, pop : { $sum : "$pop" } } },
[06:23:31] <joannac> but fine, group uniquely on (date, name, score), and grab a single instance of opp_score
[06:24:34] <Synt4x`> ok but that's what we're doing now, we have only 1 instance of opp_score for each date, i.e. {date, {name, score}, {name, score}, ..., opp_score}, but in reality there could be 50 different opp_scores that day, not just one
[06:25:13] <joannac> for every unique (date, name, score) document, you should get a single opp_score
[06:26:55] <Synt4x`> yea, I guess it is, I mean it just seems like layers of queries, I should be able to understand it I feel like, but yea, obviously I don't
[06:27:02] <Synt4x`> {'$group': { '_id': "$date", 'opponent': { '$addToSet': { 'name': "$opponent", 'score': "$score" } }, 'opp_score':{'$first':'$our_score'} }}
[06:27:10] <Synt4x`> that's what we have right now, with what you describe above, how will I change this
[06:28:29] <joannac> so that's your _id (remember that $group stage groups on whatever you tell it the _id is)
[06:31:15] <joannac> so then it looks like: $group: {_id: {date: $date, name: $name, score: $score}, opp_score: {$first: $opp_score}}
[06:31:33] <Synt4x`> '$group': { '_id': {'date':"$date", 'name': "$opponent", 'score': "$score" }, 'opp_score':{'$first':'$our_score'} }}
[06:36:21] <Synt4x`> I have it so it's filtered out the games I don't want, I need to make sure opp_score is included correctly
[06:36:25] <Synt4x`> dataSet = db.hu_db.aggregate({'$group': { '_id': {'date':"$date", 'name': "$opponent", 'score': "$score" }, 'opp_score':{'$first':'$our_score'} }, '$group': { '_id': "$date", 'opponent': { '$addToSet': { 'name': "$opponent", 'score': "$score" } }, 'opp_score':{'$first':'$our_score'} }})
[06:37:42] <Synt4x`> yes, it ran, gave me my results, but I just checked and again only 1 opp_score is at the end, so I'm assuming I made a mistake in my second query, maybe because I kept opp_score outside again the second time
[06:44:33] <Synt4x`> ok, so it's definitely my second query that is wrong, the first contains everything I would want it to with a structure of {id: {date, name, score}, opp_score}
[06:45:19] <Synt4x`> in my second I want to turn that into: {date, _id: {name, score}, oppscore:{oppscore}} **I think**
[06:48:24] <Synt4x`> originally yes, but not for the first grouping you said I just used it for the second grouping
[06:50:05] <Synt4x`> hrmm, right now my expected output is looking like this: {date, opp_score, [{name, score}, {name, score}...]}
[06:54:41] <Synt4x`> dataSet = db.hu_db.aggregate({'$group': { '_id': {'date':"$date", 'name': "$opponent", 'score': "$score" }, 'opp_score':{'$first':'$our_score'} }, '$group': { '_id': "$date", 'opponent': { '$addToSet': { 'name': "$opponent", 'score': "$score", 'opp_score':'$our_score' } } }})
[06:55:28] <Synt4x`> I tried doing 'opp_score':'$opp_score' which is what I thought it would be, but it didn't put any opp_score anywhere in the output, so I switched it to '$our_score' and I think doing that makes it read from the original set defeating the purpose
[06:55:53] <Synt4x`> so I guess the question is, how do I access opp_score from the first $group query, in the 2nd group query
[07:02:02] <Synt4x`> nevermind that failed ;(, I tried accessing like in the example, in the second group statement doing $_id.score, etc.
[07:02:13] <Synt4x`> dataSet = db.hu_db.aggregate({'$group': { '_id': {'date':"$date", 'name': "$opponent", 'score': "$score" }, 'opp_score':{'$first':'$our_score'} }, '$group': { '_id': "$date", 'opponent': { '$addToSet': { 'name': "$_id.opponent", 'score': "$_id.score", 'opp_score':'$_id.opp_score' } } }})
[07:05:33] <[1]zwoop> Im using a rest api as the only means to communicate with mongodb. The problem im haivng is that i have to specify the datamodel for every object i want to send/retrive from the database. How is this done more genral so that i can send retrive any object throgh the api?
[07:06:10] <Synt4x`> I still think I want this as expected output: {_id: date , opponent: [ {{name, score}, {opp_score}, {{name, score}, {opp_score}... ] } **or {name, score, opp_score}, {name, score, opp_score}... but I can't make either work
[07:07:37] <joannac> do you know what you are getting out of the first $group, what your documents look like?
[07:10:37] <Synt4x`> yes, the first one is doing this [ { _id: {date, score, name}, opp_score}. {_id: {date, score, name}, opp_score}...
[07:13:20] <Synt4x`> ok in the first one it's already filtered out my multiple entries, so for instance there is an entry in the db, {name:gfx_644, score: 110.24, opp_score:106.60} and then another one {name:gfx_644, score: 110.24, opp_score:94.94} (2 entries, same score, diff. opp_score)
[07:14:22] <Synt4x`> so now I'd like it to be {_id: date, {name, score, opp_score}} where all {name, score, opp_score} are already unique I believe because of the first query, so just to separate them by date and that's it
[07:18:54] <Synt4x`> sorry your right, I was thinking it auto grouped them by date but it's because I as using addToSet, so I want them to look like this: {_id : date , opponent : [ {name, score, opp_score}, {name, score, opp_score}... ] }
[07:25:40] <Synt4x`> the thing is, it's easy to see how I access the $_id.X's , how do I access opp_score from the first grouping? that's what I don't see, the example says just $pop but that didn't seem to work as I expected
[07:29:44] <Synt4x`> '$group': { '_id': "$_id.date", 'opponent': { '$addToSet': { 'name': "$_id.name", 'score': "$_id.score", 'opp_score':'$opp_score' } } }
[07:30:57] <Synt4x`> however that gives me the following: {u'ok': 1.0, u'result': [{u'_id': None, u'opponent': [{}]}]}
[07:35:36] <Synt4x`> I took out everything from the second query but this: '$group': { '_id': "$_id.date"}
[07:36:00] <Synt4x`> I am sure the 1st query still works, so I know I am getting into the second query with our normal data set
[07:38:27] <Synt4x`> dataSet = db.hu_db.aggregate({'$group': { '_id': {'date':"$date", 'name': "$opponent", 'score': "$score" }, 'opp_score':{'$first':'$our_score'} }})
[07:39:18] <Synt4x`> {u'ok': 1.0, u'result': [{u'_id': {u'date': u'2014/09/14', u'score': u'91.24', u'name': u'spacejam23'}, u'opp_score': u'92.96'}, {u'_id': {u'date': u'2014/09/14', u'score': u'118.58', u'name': u'matt201490'}, u'opp_score': u'92.96'}, ... ]}'
[07:40:49] <Synt4x`> '$group': { '_id': "$_id.date", 'opponent': { '$addToSet': { 'name': "$_id.name", 'score': "$_id.score", 'opp_score':'$opp_score' } } }
[07:42:15] <Synt4x`> dataSet = db.hu_db.aggregate({'$group': { '_id': {'date':"$date", 'name': "$opponent", 'score': "$score" }, 'opp_score':{'$first':'$our_score'} }, '$group': { '_id': "$_id.date", 'opponent': { '$addToSet': { 'name': "$_id.name", 'score': "$_id.score", 'opp_score':'$opp_score' } } }})
[07:42:37] <joannac> sorry, what's the full aggregation command with just the _id part of the second one?
[07:43:21] <Synt4x`> dataSet = db.hu_db.aggregate({'$group': { '_id': {'date':"$date", 'name': "$opponent", 'score': "$score" }, 'opp_score':{'$first':'$our_score'} }, '$group': { '_id': "$_id.date"}})
[07:44:28] <Synt4x`> 12:43am and I'm on 5 hours of sleep, yes ;(. I spent 12 hours today trying to fix this one query, something that seemed soooo simple, and I could do iteratively
[07:46:02] <Synt4x`> well I thought I figured it out, I thought I didn't have a { } around the second group
[07:46:22] <Synt4x`> and yes, you did have to prod my quite a bit, I can't tell you how thankful I am for the help already
[07:48:01] <joannac> as a hint, it should look like db.coll.aggregate({$group : {...}}, {$group: { .. }})
[07:50:44] <Synt4x`> ok wow, that was painful. Now it is hitting me with this though: TypeError: aggregate() takes exactly 2 arguments (3 given)
[07:53:37] <joannac> the mongo shell just takes multiple documents, but python needs it to be in an array
[08:06:03] <Synt4x`> Hooooooooooooooorayyyyyyyyyyyyyyyyyyyyy, got it. and MAN... what a challenging day, thanks so much to everybody here I NEED to sleep but I will pay it forward in any way possible
[09:50:43] <Forest> Hello. i am using node js with mongoDB, i am chunking array and trying to insert the JSON objects into mongodb,but not all of them are inserted,what might be the problem? I get no error message.
[09:53:42] <Forest> This is my code https://dpaste.de/NnZ2 Only 999 elemetns of cesty is isnerted and 2600 of vrcholy. Can anyone help me please?
[10:10:41] <thib> i use mongodb on a vagrant virtual machine, i've tried to update it to 2.6.5 but it seems to failed configurate mongodb-org-server, any idea why ?
[12:29:38] <ssarah> Can you tell me the easiest way to ensure external connections to a mongo machine use ssl? Or point me to the guide manual on it?
[12:49:21] <beebeeep> hello folks, does anybody knows, what means this record in config.locks: config(obj: 237636; size: 0/0/0 Gb)> db.locks.find({state: {$ne: 0}})
[13:22:05] <Forest> Hello, i am using mongodb with node js and i have a problem that array of documents i want to insert is larger than 16 MB. Can you help me how can i split the array correctly? I tried to split the array into smaller arrays of 8000 elements,but that does not guarantee the size. Any help would be appreciated.
[15:17:37] <kakashiAL> it say: give mySchema object a method that get 2 paramters, an id and a callback(cb)
[16:05:02] <hydrajump> hi if I don't specify fork=true in the mongd.conf does that mean that service start mongod will not run in the backgroun?
[17:41:32] <locojay> hi i ve removed about 100g from gridfs but when i do showdbs the db still hast a size not reflecting the delete docs
[18:39:14] <skaag> I ran rs.printReplicationInfo() and I see the last oplog event is from sometime in february 2014
[18:48:48] <skaag> now I get: Exception while invoking method 'login' MongoError: not master and slaveOk=false
[19:00:59] <hydrajump> why when I start with sudo service mongod start will it not go to the background, e.g.
[19:01:03] <hydrajump> Starting mongod: about to fork child process, waiting until server is ready for connections.
[19:26:57] <hydrajump> Derick: I need a quick check because I don't know if I'm doing the right thing
[19:27:35] <hydrajump> Derick: I have 3 members in a replicaset. I created a new member and just added it to the replicaset on the primary using rs.add("xxx")
[20:59:10] <HairAndBeardGuy_> if you're having problems doing it with mongo-hadoop, could you not use another library?
[21:30:30] <mongoexplore> cheeser, the question i asked was: I am using mongo-hadoop to convert existing json files into bson... but when I do it everything get assign as _id : {full json}
[21:35:41] <Bumptious> is there an easy way to find a gap in series of integers, each integer from a separate document?
[21:37:02] <mongoexplore> so input is {"_id":"12", "key":"field"} output bson will look like {"_id": "{\"_id":\"12\", \"key\":"\field\"} "}
[21:38:48] <cheeser> mongoexplore: i'm guessing it's because RecordWriters expect a <K,V> pair and you're only giving it one item so it assumed a null value.
[21:39:50] <Bumptious> joannac: I will be sorting them, but sometimes one will go missing and I won't know which. it's a small number, 1 to 18
[21:40:19] <mongoexplore> I have single cluster mongo server, which hold around 50gb data, when I load json it takes around 8 hours
[21:43:33] <Bumptious> i'd like to replace it. all it's doing actually would be giving "order:7" to another document
[21:46:01] <Bumptious> hmm yeah. i'm not sure why i feel compelled to make it work without that interaction
[21:48:41] <Bumptious> i think because i'm dealing with a game where there's potentially a lot of ansychronous actions going on
[21:50:50] <Bumptious> i'm just going to try it the way you suggest and see if problems arise :), thanks, joannac
[22:57:37] <mango_> Just working on a MongoDB puzzle, if there is a network partition, between D.C A (2 x secondarys) and D.C. B (1 x Primary) when the network partition has been removed, is there a re-election?
[23:00:53] <cheeser> i don't think so. that 3rd machine would just reestablish communication with the other 2 (one of which would be primary)
[23:14:42] <diegoaguilar> Hello, Im a bit confused about how I should setup MongoDB security administration
[23:15:03] <diegoaguilar> I want to create users to perform auth, have a root administrator, that's it .. all privileges
[23:15:34] <diegoaguilar> and have a some users for read access to certain databases and some users for total control of SOME databases
[23:16:05] <diegoaguilar> but at the start of docs it says that each user's role works for a unique database
[23:23:06] <mongoexplorer> have you guys use https://github.com/mongodb/mongo-hadoop/blob/master/pig/README.md
[23:24:03] <mongoexplorer> STORE raw_out INTO 'file:///tmp/whatever.bson' USING com.mongodb.hadoop.pig.BSONStorage;
[23:32:52] <mongoexplorer> REGISTER $lib_dir/mongo-hadoop-core-1.3.0.jar; REGISTER $lib_dir/mongo-hadoop-pig-1.3.0.jar;
[23:48:47] <cheeser> if there's nothing from mongo-hadoop in the stack trace, it's obviously not a mongo-hadoop problem.
[23:49:11] <cheeser> make sure the paths to your jars are correct and that you register them at the top of the script for each script you're running.