Question

我试图在MongoDB中将两个聚合列加在一起。

以下是一份示例文件：

{
    "_id" : ObjectId("52de8f56e4b0a491abb540e2"),
    "type" : "build",
    "time" : ISODate("2014-01-21T15:16:38.384Z"),
    "data": {
        "unitTests": {
            "testType": "TestNG",
            "totalTests": 6153,
            "failedTests": 2,
            "skippedTests": 1
        }
    }
}

我想在一起添加failedTests和skippedTests。这是我到目前为止的查询：

db.builds.aggregate([    
    { $match: { "data.unitTests.testType" : { $ne : null} }},
    { $group: {          
        _id: {              
            month: { $month: "$time" },             
            day: { $dayOfMonth: "$time" },             
            year: { $year: "$time" },                      
        }, 
        total: { $avg: "$data.unitTests.totalTests" },
        failed: { $avg: "$data.unitTests.failedTests" },
        skipped: { $avg: "$data.unitTests.skippedTests" },  
        error: { $add: ["$data.unitTests.failedTests", "$data.unitTests.skippedTests"] },
    } },     
    { $sort: { "_id.year": 1, "_id.month": 1, "_id.day": 1 } } 
])

错误元素是我试图将两者加在一起的那个 - 我已经尝试了其他几种语法选项而没有太大的成功。谢谢。

Answer 1

您应该将$add与$project一起使用，而不是$group。

db.builds.aggregate([    
{ $match: { "data.unitTests.testType" : { $ne : null} }},
{ $group: {          
    _id: {              
        month: { $month: "$time" },             
        day: { $dayOfMonth: "$time" },             
        year: { $year: "$time" },                      
    }, 
    total: { $avg: "$data.unitTests.totalTests" },
    failed: { $avg: "$data.unitTests.failedTests" },
    skipped: { $avg: "$data.unitTests.skippedTests" }

} },
{ $project: { 
    total: 1,
    failed: 1,
    skipped: 1,
    error: { $add: ["$failed", "$skipped"] }
} },   
{ $sort: { "_id.year": 1, "_id.month": 1, "_id.day": 1 } } 
])

Answer 2

您可以使用管道中的$ project阶段来执行此操作，例如：

db.builds.aggregate([     
{$match: {"data.unitTests.testType": {$ne: null}}},
{$group: {_id: {month: {$month: "$time"},day: {$dayOfMonth: "$time"},year: {$year: "$time"}},
        total: {$avg: "$data.unitTests.totalTests"},
        failed: {$avg: "$data.unitTests.failedTests"},
        skipped: {$avg: "$data.unitTests.skippedTests"}}
},
{$project: {total: 1,
        failed: 1,
        skipped: 1,
        error: {$add: ["$failed","$skipped"]}}
},
{$sort: {"_id.year": 1,"_id.month": 1,"_id.day": 1}}
])

Answer 3

尝试第二个group

db.builds.aggregate([    
    { $match: { "data.unitTests.testType" : { $ne : null} }},
    { $group: {          
        _id: {              
            month: { $month: "$time" },             
            day: { $dayOfMonth: "$time" },             
            year: { $year: "$time" },                      
        }, 
        total: { $avg: "$data.unitTests.totalTests" },
        failed: { $avg: "$data.unitTests.failedTests" },
        skipped: { $avg: "$data.unitTests.skippedTests" }
    } },
    { $group: { _id: '$_id', total: '$total', failed: '$failed', skipped: '$skipped', error: { $add: ['$failed', '$skipped'] } },     
    { $sort: { "_id.year": 1, "_id.month": 1, "_id.day": 1 } } 
])

MongoDB添加两个聚合列

3 个答案: