【发布时间】:2022-01-21 20:06:24
【问题描述】:
我在 MongoDB 5 中有一个名为“漏洞”的集合。它看起来像这样:
[
{ _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 }
]
我想为每个扫描仪检索具有最高版本的文档。它应该返回:
[
{ _id: "...", "project": 244, "scanner": "sonarqube", "version": 2 },
{ _id: "...", "project": 244, "scanner": "sonarqube", "version": 2 },
{ _id: "...", "project": 244, "scanner": "shellcheck", "version": 3 },
{ _id: "...", "project": 244, "scanner": "powershell", "version": 4 },
{ _id: "...", "project": 244, "scanner": "powershell", "version": 4 }
]
我想只用一个查询来完成。
有什么建议吗?
非常感谢
【问题讨论】:
标签: database mongodb collections max aggregation