mongodb - How to exclude _id without including other fields using the aggregation framework -


i result of aggregation pipeline without _id filed. know possible if provide explicitly other fields output of projection. but, ¿how can mimic $projec in find call?

this want (no field explicitly included):

db.col.find({},{_id:0})

but in aggregation framework seems impossible:

db.col.aggregate([{'$project': {_id:0}}])  error: printing stack trace     @ printstacktrace (src/mongo/shell/utils.js:37:15)     @ dbcollection.aggregate (src/mongo/shell/collection.js:927:9)     @ (shell):1:11 2013-10-07t16:36:09.273+0200 aggregate failed: {     "errmsg" : "exception: $projection requires @ least 1 output field",     "code" : 16403,     "ok" : 0 } @ src/mongo/shell/collection.js:928

any idea workaround issue ?

when using aggregation, must explicitly include/exclude fields. so, you'd need list fields want. it's not equivalent find. so, might:

db.sample.aggregate(     { $project : {         _id : 0,         title : 1                  }} );

using aggregation framework comes limits should aware of. it's designed aggregation (grouping, summing, etc.), having many fields in projection isn't typical (and cause results exceed maximum allowed, 16mb).


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