0

我在 MongoDB 中有这个集合：

[
  { _id: "...", "project": 244, "scanner": "sonarqube", "version": 1 },
  { _id: "...", "project": 244, "scanner": "sonarqube", "version": 2 },
  { _id: "...", "project": 244, "scanner": "sonarqube", "version": 2 },
  { _id: "...", "project": 244, "scanner": "shellcheck", "version": 1 },
  { _id: "...", "project": 244, "scanner": "shellcheck", "version": 2 },
  { _id: "...", "project": 244, "scanner": "shellcheck", "version": 3 },
  { _id: "...", "project": 244, "scanner": "powershell", "version": 2 },
  { _id: "...", "project": 244, "scanner": "powershell", "version": 3 },
  { _id: "...", "project": 244, "scanner": "powershell", "version": 4 },
  { _id: "...", "project": 244, "scanner": "powershell", "version": 4 }
]

我想返回按扫描仪分组的最新（最高）版本的项目数：

{ scanner: "sonarqube", count: 2 }, // All items with sonarqube and version: 2
{ scanner: "shellcheck", count: 1 }, // All items with shellcheck and version: 3
{ scanner: "powershell", count: 2 } // All items with powershell and version: 4

到目前为止，我得出了这个：

db.getCollection("vulnerabilities").aggregate([
{
    $match: { project_id: 422 }
},
{
    $sort: { version: -1 }
},
{
    $group: {
        _id: "$scanner",
        'count': { $first: "$version"},
    }
}
])

这将返回每个组的最新（最高）版本：

sonarqube  | 2 |
shellcheck | 3 |
powershell | 4 |

现在，我需要告诉 Mongo：

  • 使用该版本过滤项目
  • 按扫描仪分组
  • 添加计数

有什么想法、建议吗？

谢谢

4

2 回答 2

1

最简单的方法是在按版本排序之前先获取计数，也许：

db.getCollection("vulnerabilities").aggregate([
{
    $match: { project_id: 422 }
},
{   
    $group:{
       _id: { scanner: "$scanner", version: "$version"},
       count: { $sum: 1 }
    }
},
{
    $sort: { "_id.version": -1 }
},
{
    $group: {
        _id: "$_id.scanner",
        count: { $first: "$count" },
        version: { $first: "$_id.version" }
    }
}
])
于 2022-02-03T23:47:59.460 回答
0

你可以$max在$group舞台上使用

db.test.aggregate([ 
    {$group : {_id : "$scanner", count : {$sum : 1}, version : {$max : "$version"}}}
])

或者您可以$sort按版本和使用$first

db.test.aggregate([
    {$sort : {version: -1}}, 
    {$group : {_id : "$scanner", count : {$sum : 1}, version : {$first : "$version"}}}
])
于 2022-02-04T03:23:41.987 回答