Question

我仍然是MongoDB的新手，我正试图弄清楚当其中一个是计算日期字段时如何按多个字段分组：

我的数据看起来像这样（简化）：

{
    "_id" : ObjectId("52d6ed19e4b0a491abb53ff2"),
    "build_duration" : 667075,
    "build_number" : 40,
    "build_result" : "SUCCESS",
    "build_timestamp" : ISODate("2014-01-15T20:07:27.324Z"),
    "job_name" : "ABC"
}

我知道我可以通过多个字段进行分组：

db.builds.aggregate(
   { $group: { _id: { build_result: "$build_result", job_name: "$job_name"},
           build_duration: { $avg: "$build_duration" } } }
)

但是，如果我尝试使用$ dayOfYear函数对日期进行分组，则会出现错误：

db.builds.aggregate(
   { $group: { _id: { build_result: "$build_result", $dayOfYear: "$build_timestamp"},
           build_duration: { $avg: "$build_duration" } } }
)

错误：

Error: Printing Stack Trace
    at printStackTrace (src/mongo/shell/utils.js:37:15)
    at DBCollection.aggregate (src/mongo/shell/collection.js:897:9)
    at (shell):1:11
Thu Jan 16 13:34:08.036 aggregate failed: {
    "errmsg" : "exception: the operator must be the only field in a pipeline object (at '$dayOfYear'",
    "code" : 15983,
    "ok" : 0
} at src/mongo/shell/collection.js:898

所需的输出是按日期分组，build_result是build_duration的平均值。提前谢谢。

Answer 1

您需要使用$dayOfYear对象作为_id中新字段的值，如下所示：

db.builds.aggregate(
    { $group: { 
        _id: { 
            build_result: "$build_result", 
            dayOfYear: { $dayOfYear: "$build_timestamp" }
        },
        build_duration: { $avg: "$build_duration" } 
    } }
)

输出：

{
  "result": [
    {
      "_id": {
        "build_result": "SUCCESS",
        "dayOfYear": 15
      },
      "build_duration": 667075
    }
  ],
  "ok": 1
}

当一个是日期字段时，如何在MongoDB中按多个字段进行分组

1 个答案: