Question

我们正在使用mongoose

我们有这个集合

{
    "_id" : ObjectId("5596c195336cbb042728c83b"),
    "cameraId" : ObjectId("559514d3f20fb7e959a5b078"),
    "orignalFilePath" : "/opt/safecamz/recording/55950adaa24f46d255acfe90/559514d3f20fb7e959a5b078/1A01270PAA0012/2000-01-09/001/dav/23/23.26.09-23.27.33[M][0@0][0].dav",
    "recordingDuration" : 11.042,
    "recordingLocation" : "/opt/safecamz/recording/55950adaa24f46d255acfe90/559514d3f20fb7e959a5b078/video/23.26.09-23.27.33[M][0@0][0].mp4",
    "userId" : ObjectId("55950adaa24f46d255acfe90"),
    "created" : ISODate("2015-07-03T17:08:37.537Z"),
    "recordingSize" : "1259.3857421875",
    "recordingWowzaUrl" : "55950adaa24f46d255acfe90/559514d3f20fb7e959a5b078/video/23.26.09-23.27.33[M][0@0][0].mp4",
    "recordingName" : "23.26.09-23.27.33[M][0@0][0].mp4",
    "__v" : 0
}

现在我们要通过 userId 添加 recordingSize 例如，如果有10,000个记录，那么recordingSize的总和，其中userId = ObjectId（＆＃34; 55950adaa24f46d255acfe90＆＃34;）

我们尝试使用$sum，但我们无法获得正确的输出

Answer 1

您可以使用聚合。让我们假设样本数据来自您的样本：

> db.recordings.find({}, {_id: 0, userId: 1, recordingSize: 1, recordingDuration:1})
{ "recordingDuration" : 11.042, "userId" : ObjectId("55950adaa24f46d255acfe90"), "recordingSize" : "1259.3857421875" }
{ "recordingDuration" : 11.042, "userId" : ObjectId("55950adaa24f46d255acfe90"), "recordingSize" : "3000.3857421875" }

让我们先试试recordingDuration然后看totalDuration中的总和：

> db.recordings.aggregate([{$project: {userId:1, recordingDuration:1}}, {$group: { _id: '$userId', totalDuration: { $sum: '$recordingDuration' }}}])
{ "_id" : ObjectId("55950adaa24f46d255acfe90"), "totalDuration" : 22.084 }

问题是recordingSize是一个字符串，这就是为什么$sum无效的原因：

> db.recordings.aggregate([{$project: {userId:1, recordingSize:1}}, {$group: { _id: '$userId', totalSize: { $sum: '$recordingSize' }}}])
{ "_id" : ObjectId("55950adaa24f46d255acfe90"), "totalSize" : 0 }

此处有更多详情：$sum (aggregation)。

通过匹配特定字段来匹配特定字段的总和

1 个答案: