聚合与条件和分组匹配的子文档的数量

Question

我收集的文件如下：

{
    "_id" : ObjectId("55d4410544c96d6f6578f893"),
    "executionProject" : "Project1",
    "suiteList" : [ 
        {
            "suiteStatus" : "PASS"
        }
    ],
    "runEndTime" : ISODate("2015-08-19T08:40:47.049Z"),
    "runStartTime" : ISODate("2015-08-19T08:40:37.621Z"),
    "runStatus" : "PASS",
    "__v" : 1
}

{
    "_id" : ObjectId("55d44eb4c0422e7b8bffe76b"),
    "executionProject" : "Project1",
    "suiteList" : [ 
        {
            "suiteStatus" : "PASS"
        }
    ],
    "runEndTime" : ISODate("2015-08-19T09:39:13.528Z"),
    "runStartTime" : ISODate("2015-08-19T09:39:00.406Z"),
    "runStatus" : "PASS",
    "__v" : 1
}

{
    "_id" : ObjectId("55d44f0bc0422e7b8bffe76f"),
    "executionProject" : "Project1",
    "suiteList" : [ 
        {
            "suiteStatus" : "FAIL"
        }
    ],
    "runEndTime" : ISODate("2015-08-19T09:46:31.108Z"),
    "runStartTime" : ISODate("2015-08-19T09:40:27.377Z"),
    "runStatus" : "PASS",
    "__v" : 1
}

{
    "_id" : ObjectId("55d463d0c0422e7b8bffe789"),
    "executionProject" : "Project2",
    "suiteList" : [ 
        {
            "suiteStatus" : "PASS"
        },
        {
            "suiteStatus" : "PASS"
        }
    ],
    "runEndTime" : ISODate("2015-08-19T11:09:52.537Z"),
    "runStartTime" : ISODate("2015-08-19T11:09:04.539Z"),
    "runStatus" : "FAIL",
    "__v" : 1
}

{
    "_id" : ObjectId("55d464ebc0422e7b8bffe7c2"),
    "executionProject" : "Project3",
    "suiteList" : [ 
        {
            "suiteStatus" : "FAIL"
        }
    ],
    "runEndTime" : ISODate("2015-08-19T11:18:41.460Z"),
    "runStartTime" : ISODate("2015-08-19T11:13:47.268Z"),
    "runStatus" : "FAIL",
    "__v" : 10
} 

我期望输出如下：

[
    {
        "executionProject": "Project1",
        "suite-pass": 0,
        "suite-fail": 1,
        "runEndTime": ISODate("2015-08-19T09:46:31.108Z")
    },
    {
        "executionProject": "Project2",
        "suite-pass": 2,
        "suite-fail": 0,
        "runEndTime": ISODate("2015-08-19T11:09:52.537Z")
    },
    {
        "executionProject": "Project3",
        "suite-pass": 0,
        "suite-fail": 1,
        "runEndTime": ISODate("2015-08-19T11:18:41.460Z")
    },
]

我想按项目分组并按 runEndTime 排序并显示 suiteList 的通过和失败计数。

我按照 Blakes 在Mongodb: Group by element and show the sub-document count based on condition and sort the document by date 中的建议尝试了这个：

db.testruns.aggregate([
      { "$sort": { "runEndTime": 1 } },
      { "$group": {
      "_id": "$executionProject",
        "suite-pass": {
          "$last": {
            "$cond": [
              { "$anyElementTrue": {
                "$map": {
                  "input": "$suiteList",
                  "as": "suite",
                  "in": {
                    "$eq": [ "$$suite.suiteStatus", "PASS" ]
                  }
                }
              }},
              1,
              0
            ]
          }
        },
        "suite-fail": {
          "$last": {
            "$cond": [
              { "$anyElementTrue": {
                "$map": {
                  "input": "$suiteList",
                  "as": "suite",
                  "in": {
                    "$eq": [ "$$suite.suiteStatus", "FAIL" ]
                  }
                }
              }},
              1,
              0
            ]
          }
        },
        "runEndTime": { "$last": "$runEndTime" }
      }},
      { "$sort": { "runEndTime": 1 } }
    ]);

我期待 Project2 的 suite-pass 计数为 2，因为 suiteList 中有 2 个元素，但它返回 1。

Answer 1

您应该已经正确阅读了答案，因为已经有另一个替代列表和解释为什么您想要的预期结果与您使用的结果不同。

相反，您想要这个，它尊重可能的多个“PASS”或“FAIL”：

  Model.aggregate(
    [
      { "$sort": { "executionProject": 1, "runEndTime": 1 } },
      { "$group": {
        "_id": "$executionProject",
        "suiteList": { "$last": "$suiteList" },
        "runEndTime": { "$last": "$runEndTime" }
      }},
      { "$unwind": "$suiteList" },
      { "$group": {
        "_id": "$_id",
        "suite-pass": { 
          "$sum": {
            "$cond": [
              { "$eq": [ "$suiteList.suiteStatus", "PASS" ] },
              1,
              0
            ]
          }
        },
        "suite-fail": { 
          "$sum": {
            "$cond": [
              { "$eq": [ "$suiteList.suiteStatus", "FAIL" ] },
              1,
              0
            ]
          }
        },
        "runEndTime": {"$first": "$runEndTime"}
      }},
      { "$sort": { "runEndTime": 1 }}
    ],
    function(err,result) {

    }
  );

这是一种方法的“组合”。第一个是按照您的预期通过 runTime 获得“最后一个”。接下来是分解数组，这一次实际上“总结”可能出现的通过或失败，而不是只记录数组中通过或失败的1，实际的“通过”或“失败”被计算在内。

有结果：

{
        "_id" : "Project1",
        "suite-pass" : 0,
        "suite-fail" : 1,
        "runEndTime" : ISODate("2015-08-19T09:46:31.108Z")
}
{
        "_id" : "Project2",
        "suite-pass" : 2,
        "suite-fail" : 0,
        "runEndTime" : ISODate("2015-08-19T11:09:52.537Z")
}
{
        "_id" : "Project3",
        "suite-pass" : 0,
        "suite-fail" : 1,
        "runEndTime" : ISODate("2015-08-19T11:18:41.460Z")
}

Answer 2

展开suiteList 并在group 中使用$sum，如下所示：

db.testruns.aggregate({
    "$unwind": "$suiteList"
}, {
    "$group": {
        "_id": "$executionProject",
        "suite-pass": {
            "$sum": {
                "$cond": {
                    "if": {
                        "$eq": ["$suiteList.suiteStatus", "PASS"]
                    },
                    "then": 1,
                    "else": 0
                }
            }
        },
        "suite-fail": {
            "$sum": {
                "$cond": {
                    "if": {
                        "$eq": ["$suiteList.suiteStatus", "FAIL"]
                    },
                    "then": 1,
                    "else": 0
                }
            }
        },
        "runEndTime": {
            "$last": "$runEndTime"
        }
    }
}, {
    "$sort": {
        "runEndTime": 1
    }
})