MongoDB过滤嵌套数组

huangapple
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英文:

MongoDB filter nested array

问题

  1. [
  2.   {
  3.     "_id": 1,
  4.     "title": "dummy title",
  5.     "settings": [
  6.       {
  7.         "type": "light",
  8.         "status": "enabled"
  9.       },
  10.       {
  11.         "type": "toolbar",
  12.         "status": "enabled"
  13.       }
  14.     ]
  15.   }
  16. ]

我想按ID获取它，但只想获取“enabled”设置，不确定聚合管道应该是什么样子的。

我应该使用Java / Kotlin中的mongoTemplate创建查询，但仅有Mongo查询即可。

英文:

I have a collection that looks something like:

  1. [
  2.   {
  3.     "_id": 1,
  4.     "title": "dummy title",
  5.     "settings": [
  6.       {
  7.         "type": "light",
  8.         "status": "enabled"
  9.       },
  10.       {
  11.         "type": "flare",
  12.         "status": "disabled"
  13.       },
  14.       {
  15.         "type": "toolbar",
  16.         "status": "enabled"
  17.       }
  18.     ]
  19.   }
  20. ]

I wanna fetch it by Id but only with the "enabled" settings and not sure how the aggregation pipeline should look like.

I supposed to create the query using mongoTemplate in Java / Kotlin but even just the mongo query would be enough.

答案1

得分: 1

  1. // MongoDB shell/CLI版本4.26的工作代码（在Windows 10机器/操作系统上）
  2. > db.titles.find().pretty();
  3. {
  4.         "_id" : 1,
  5.         "title" : "虚拟标题",
  6.         "settings" : [
  7.                 {
  8.                         "type" : "light",
  9.                         "status" : "已启用"
  10.                 },
  11.                 {
  12.                         "type" : "flare",
  13.                         "status" : "已禁用"
  14.                 },
  15.                 {
  16.                         "type" : "toolbar",
  17.                         "status" : "已启用"
  18.                 }
  19.         ]
  20. }
  21. > db.titles.aggregate([
  22. ... {$unwind:"$settings"},
  23. ... {$match:{"settings.status":"已启用"}}
  24. ... ]);
  25. { "_id" : 1, "title" : "虚拟标题", "settings" : { "type" : "light", "status" : "已启用" } }
  26. { "_id" : 1, "title" : "虚拟标题", "settings" : { "type" : "toolbar", "status" : "已启用" } }
  27. > print("MongoDB: ",db.version());
  28. MongoDB:  4.2.6
  29. >
英文: 
  1. //working code from MongoDB shell/CLI version 4.26(on windows 10 machine/OS)
  2. > db.titles.find().pretty();
  3. {
  4.         "_id" : 1,
  5.         "title" : "dummy title",
  6.         "settings" : [
  7.                 {
  8.                         "type" : "light",
  9.                         "status" : "enabled"
  10.                 },
  11.                 {
  12.                         "type" : "flare",
  13.                         "status" : "disabled"
  14.                 },
  15.                 {
  16.                         "type" : "toolbar",
  17.                         "status" : "enabled"
  18.                 }
  19.         ]
  20. }
  21. > db.titles.aggregate([
  22. ... {$unwind:"$settings"},
  23. ... {$match:{"settings.status":"enabled"}}
  24. ... ]);
  25. { "_id" : 1, "title" : "dummy title", "settings" : { "type" : "light", "status" : "enabled" } }
  26. { "_id" : 1, "title" : "dummy title", "settings" : { "type" : "toolbar", "status" : "enabled" } }
  27. > print("MongoDB: ",db.version());
  28. MongoDB:  4.2.6
  29. >

答案2

得分: 1

你可以使用$filter来实现。

  1. db.collection.aggregate([
  2.   {
  3.     $project: {
  4.       _id: 1,
  5.       title: 1,
  6.       settings: {
  7.         $filter: {
  8.           input: "$settings",
  9.           as: "item",
  10.           cond: {
  11.             $eq: [
  12.               "$$item.status",
  13.               "enabled"
  14.             ]
  15.           }
  16.         }
  17.       }
  18.     }
  19.   }
  20. ])

在此链接中尝试。

英文:

You can do it with a $filter

  1. db.collection.aggregate([
  2.   {
  3.     $project: {
  4.       _id: 1,
  5.       title: 1,
  6.       settings: {
  7.         $filter: {
  8.           input: "$settings",
  9.           as: "item",
  10.           cond: {
  11.             $eq: [
  12.               "$$item.status",
  13.               "enabled"
  14.             ]
  15.           }
  16.         }
  17.       }
  18.     }
  19.   }
  20. ])

try it here

huangapple
  • 本文由 发表于 2020年10月12日 00:03:31
  • 转载请务必保留本文链接：https://go.coder-hub.com/64306166.html
  • java
  • kotlin
  • mongodb
  • mongotemplate
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