mongodb 字段“$name”必须是累加器对象
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The field "$name" must be an accumulator object
提问by Matheus Barem
I have a query and when I use a $groupa error shows "the field "$name must be an accumulator object", if if remove the filed "$name" all works well and i have tried to use only "name" instead of "$name" and the error continues.
我有一个查询，当我使用$group 时，错误显示“字段“$name 必须是累加器对象”，如果删除提交的“$name”一切正常并且我尝试仅使用“name” "$name" 并且错误继续。
User.aggregate([
{
$match: {
"storeKey": req.body.store
}
},
{
$group: {
"_id": "$_id",
"name": "$name",
"count": {
"$sum": 1
},
"totalValue": {
"$sum": "$value"
}
}
},
{
$sort: sort
},
{
$skip: req.body.limit * req.body.page
},
{
$limit: req.body.limit
}
])...
回答by Ashh
There are some aggregation operators that can only be used in $groupaggregation and named as $group accumulators
有一些聚合运算符只能在$group聚合中使用并命名为$group accumulators
Just as you used $sumhere you have to use for the namekey as well
就像您$sum在这里使用的一样，您也必须将其用于name密钥
{ "$group": {
"_id": "$_id",
"name": { "$first": "$name" }, //$first accumulator
"count": { "$sum": 1 }, //$sum accumulator
"totalValue": { "$sum": "$value" } //$sum accumulator
}}
Accumulator is like array of Elements its Accumulates as Array. $first -> gives 1st name that goes in the group of names
累加器就像元素数组，它以数组形式累加。$first -> 给出名称组中的第一个名称
Example:
so if you have $_idsame but different name ["Darik","John"]specifying $firstwill give Darik & similarly $lastwill give John
示例：因此，如果您具有$_id相同但不同的名称，则["Darik","John"]指定$first将给 Darik，同样$last将给 John
回答by Manish Solanki
db.faq_feedback.aggregate({
$lookup:{
"from":"faq",
"localField":"question_id",
"foreignField":"_id",
"as":"faq"
}
},
{
$unwind:"$faq"
},
{
$project:{
"question_id":1,
"lang":"$faq.lang",
"feedback":"$faq.feedback",
"question":"$faq.question",
"yes":{
"$cond":[
{
"$eq":[
"$feedback",
"yes"
]
},
1,
0
]
},
"no":{
"$cond":[
{
"$eq":[
"$feedback",
"no"
]
},
1,
0
]
}
}
},
{
$group:{
"_id":"$question_id",
"yes":{
"$sum":"$yes"
},
"no":{
"$sum":"$no"
},
"question":{"$first":"$question"},
"lang":{"$first":"$lang"}
}
},
{
$limit:10000
},
{
$skip:0
})
回答by Mohit Kumar Bordia
$group: {
_id:"$_id",
"name" :{ $last: '$name' },
"count" : { $sum: 1 },
"totalValue": { "$sum": "$value" }
}
