MongoDB 按组中检索到的值过滤
MongoDB filter by retrieved value in group
我在 MongoDB 中有这个 collection:
[
{ _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 }
]
我想要 return 项目数,按扫描仪分组,最新(最高)版本:
{ 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"},
}
}
])
这是每个组的 return 最新(最高)版本:
sonarqube | 2 |
shellcheck | 3 |
powershell | 4 |
现在,我要告诉Mongo:
- 使用该版本筛选项目
- 按扫描仪分组
- 添加计数
有什么想法、建议吗?
谢谢
最简单的方法是先获取计数,然后再按版本排序,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" }
}
}
])
在$group
阶段可以使用$max
db.test.aggregate([
{$group : {_id : "$scanner", count : {$sum : 1}, version : {$max : "$version"}}}
])
或者您可以 $sort
按版本并使用 $first
db.test.aggregate([
{$sort : {version: -1}},
{$group : {_id : "$scanner", count : {$sum : 1}, version : {$first : "$version"}}}
])
我在 MongoDB 中有这个 collection:
[
{ _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 }
]
我想要 return 项目数,按扫描仪分组,最新(最高)版本:
{ 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"},
}
}
])
这是每个组的 return 最新(最高)版本:
sonarqube | 2 |
shellcheck | 3 |
powershell | 4 |
现在,我要告诉Mongo:
- 使用该版本筛选项目
- 按扫描仪分组
- 添加计数
有什么想法、建议吗?
谢谢
最简单的方法是先获取计数,然后再按版本排序,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" }
}
}
])
在$group
阶段可以使用$max
db.test.aggregate([
{$group : {_id : "$scanner", count : {$sum : 1}, version : {$max : "$version"}}}
])
或者您可以 $sort
按版本并使用 $first
db.test.aggregate([
{$sort : {version: -1}},
{$group : {_id : "$scanner", count : {$sum : 1}, version : {$first : "$version"}}}
])