Question

例如，我有这些数据：

{project: "1": platform: "1", number: 10}
{project: "1": platform: "1", number: 10}
{project: "1": platform: "1", number: 40}
{project: "1": platform: "1", number: 40}

{project: "1": platform: "2", number: 20}
{project: "1": platform: "2", number: 20}
{project: "1": platform: "2", number: 30}
{project: "1": platform: "2", number: 30}

{project: "2": platform: "2", number: 50}
{project: "2": platform: "2", number: 50}
{project: "2": platform: "2", number: 60}
{project: "2": platform: "2", number: 60}

我想获得按项目和平台分组的行，并获取具有最大数量的所有行。上述数据的结果应为：

{project: "1": platform: "1", number: 40}
{project: "1": platform: "1", number: 40}
{project: "1": platform: "2", number: 30}
{project: "1": platform: "2", number: 30}
{project: "2": platform: "2", number: 60}
{project: "2": platform: "2", number: 60}

我尝试使用$ max在$ group._id中使用项目和平台进行聚合，但查询只返回一行最大数量。如何通过mongodb制作？

Answer 1

以下是基于$filter运算符的解决方案：

db.projects.aggregate([{$group: {_id: {project: "$project", platform: "$platform"}, numbers: {$push: "$number"}, max: {$max: "$number"}}},
                       {$project: {_id: 0,
                                   project: "$_id.project",
                                   platform: "$_id.platform",
                                   number: {$filter: {
                                                      input: "$numbers",
                                                      as: "number",
                                                      cond: {$eq: ["$$number", "$max"]}
                                                     }
                                           }
                                  }
                       },
                       {$unwind: "$number"}]);

如果原始文档中有更多字段，您可以尝试：

As far as I understand you need something like this db.projects.aggregate([{$group: {_id: {project: "$project", platform: "$platform"}, documents: {$push: "$$ROOT"}, max: {$max: "$number"}}},
                       {$project: {_id: 0,
                                   document: {$filter: {
                                                      input: "$documents",
                                                      as: "document",
                                                      cond: {$eq: ["$$document.number", "$max"]}
                                                     }
                                           }
                                  }
                       },
                       {$unwind: "$document"},
                       {$project: {_id: "$document._id", number: "$document.number", ANOTHER_FIELD: "$ANOTHER_FIELD"}}]);

Answer 2

在进入聚合查询之前检查以下链接以供参考：

1＆GT;的 Encode method source

2 - ;的 Mongo $eq

3＆GT;的 $setDifference

4＆GT;的 $sond

首先我们按platform和project进行分组，然后在聚合中使用number查找最大group，并在组中使用$$ROOT添加所有数据。然后使用$map迭代所有数据并检查匹配的最大数量，并仅获取那些最大匹配数据。

db.collection.aggregate([{ "$group": { "_id": { "project": "$project", "paltform": "$platform" }, "max": { "$max": "$number" }, "mainData": { "$push": "$$ROOT" } } }, { "$project": { "findMax": { "$setDifference": [{ "$map": { "input": "$mainData", "as": "el", "in": { "$cond": { "if": { "$eq": ["$$el.number", "$max"] }, "then": "$$el", "else": false } } } }, [false] ] }, "_id": 0 } }, { "$unwind": "$findMax" }, { "$project": { "_id": "$findMax._id", "project": "$findMax.project", "platform": "$findMax.platform", "number": "$findMax.number" } }])

如果您删除了上一个uniwnd和project，也可以获得结果。

获取所有行，分组和最大值

2 个答案: