In my collection there are documents like this
{ "_id" : 112, "name" : "Myrtle Wolfinger", "scores" : [ { "type" : "exam", "score" : 73.93895528856032 }, { "type" : "quiz", "score" : 35.99397009906073 }, { "type" : "homework", "score" : 93.85826506506328 }, { "type" : "homework", "score" : 71.21962876453497 } ] }
I want to find for each document the min of the field scores.score where score.type = "homework"
.
I executed a query like this
db.students.find({},{"scores.score":1}).min( { "scores.type":"homework" } )
mongo shell returns this error
error: {
"$err" : "Unable to execute query: error processing query: ns=school.students limit=0 skip=0\nTree: $and\nSort: {}\nProj: { scores.score: 1.0 }\n planner returned error: unable to find relevant index for max/min query",
"code" : 17007
}
You wouldn't use find() with min() for this. min() would simple limit your results to those above a lower bound using an index. Instead try using aggregate().
db.students.aggregate([
{
$unwind:"$scores"
},
{
$match: {"scores.type":"homework"}
},
{
$group:{
_id: {
_id:"$_id",
name: "$name"
},
min_score: {$min: "$scores.score"}
}
}
]);