在MongoDB中对对象数组进行分组并获取计数。

huangapple
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英文:

group array of objects and get the count in mongodb

问题

  1. 我有一个对象数组：
  3. [
  4.   {
  5.     medicationDetails: {
  6.       category: 'Asthma',
  7.       subCategory: '',
  8.       providedBy: 'Doctor'
  9.     }
  10.   },
  11.   {
  12.     medicationDetails: {
  13.       category: 'Diabetes',
  14.       subCategory: 'Oral',
  15.       providedBy: 'Nurse'
  16.     }
  17.   },
  18.   {
  19.     medicationDetails: {
  20.       category: 'Asthma',
  21.       subCategory: '',
  22.       providedBy: 'Doctor'
  23.     }
  24.   },
  25.   {
  26.     medicationDetails: {
  27.       category: 'Diabetes',
  28.       subCategory: 'Insulin',
  29.       providedBy: 'Doctor'
  30.     }
  31.   }
  32. ]
  34. 一些药物可能有子类别，而其他的则没有。
  36. 需要根据'category'、'sub-category'和'providedBy'进行分组，并获得如下计数：
  38. [
  39.   {
  40.     medicationDetails: [
  41.       {
  42.         name: 'Asthma',
  43.         providedByDoctorCount: 2,
  44.         providedByNurseCount: 0
  45.       },
  46.       {
  47.         name: 'Oral',
  48.         providedByDoctorCount: 0,
  49.         providedByNurseCount: 1
  50.       },
  51.       {
  52.         name: 'Insulin',
  53.         providedByDoctorCount: 1,
  54.         providedByNurseCount: 0
  55.       }
  56.     ]
  57.   }
  58. ]
  60. 我尝试了$group和$count，但没有得到预期的结果。
  61. 需要一些有价值的帮助。
英文:

I am having an array of objects:

  1. [
  2.   {
  3.     medicationDetails: {
  4. 	  category: 'Asthma',
  5. 	  subCategory: '',
  6. 	  providedBy: 'Doctor'
  7.     }
  8.   },
  9.   {
  10.     medicationDetails: {
  11. 	  category: 'Diabetes',
  12. 	  subCategory: 'Oral',
  13. 	  providedBy: 'Nurse'
  14.     }
  15.   },
  16.   {
  17.     medicationDetails: {
  18. 	  category: 'Asthma',
  19. 	  subCategory: ''
  20. 	  providedBy: 'Doctor'
  21.     }
  22.   },
  23.   {
  24.     medicationDetails: {
  25. 	  category: 'Diabetes',
  26. 	  subCategory: 'Insulin',
  27. 	  providedBy: 'Doctor'
  28.     }
  29.   }
  30. ]

Some medication will have sub-category and some doesn't.

Need to group based on 'category', 'sub-category' and 'providedBy' and get the count like this:

  1. [
  2.   {
  3.     medicationDetails: [
  4.      {
  5.       name: 'Asthma',
  6. 	  providedByDoctorCount: 2,
  7. 	  providedByNurseCount: 0
  8.      },
  9.      {
  10.       name: 'Oral',
  11.       providedByDoctorCount: 0,
  12. 	  providedByNurseCount: 1
  13.      },
  14.      {
  15.       name: 'Insulin',
  16.       providedByDoctorCount: 1,
  17. 	  providedByNurseCount: 0
  18.      }
  19.     ]
  20.   }
  21. ]

I was tried with $group and $count but not getting the expected result.
Need some valuable help.

答案1

得分: 1

  1. 基本上与您的先前问题类似。

  2. 在分组时，可以根据字段的值进行有条件求和。

  3. 基于 category 和 subCategory 的值创建一个 key 字段。

  4. 根据 providedBy 的值，有条件地将其添加到 providedByDoctor 或 providedByNurse 中。

  5. 将其分组成单个的 medicationDetails 数组。

  1. db.collection.aggregate([
  2.   { 
  3.    $project: {
  4.       providedBy: "$medicationDetails.providedBy",
  5.       key: { $cond: [ { $eq: [ "$medicationDetails.subCategory", "" ] }, "$medicationDetails.category", "$medicationDetails.subCategory" ] }
  6.     } 
  7.   },
  8.   {
  9.     $group: { 
  10.       _id: "$key",
  11.       providedByDoctor: { $sum: { $cond: [ { $eq: [ "$providedBy", "Doctor" ] }, 1, 0 ] } },
  12.       providedByNurse: { $sum: { $cond: [ { $eq: [ "$providedBy", "Nurse" ] }, 1, 0 ] } } }
  13.   },
  14.   { $group: { _id: null, medicationDetails: { $push: { name: "$_id", providedByDoctorCount: "$providedByDoctor", providedByNurseCount: "$providedByNurse" } } } },
  15.   { $unset: "_id" }
  16. ])

playground

英文:

somewhat similar to your previous question.
you can conditionally sum based on the value of the field when grouping

  1. Create a key field based on category and subCategory values
  2. Group by the key and add to providedByDoctor or providedByNure conditionally based on the value of providedBy
  3. Group into a single medicationDetails array
  1. db.collection.aggregate([
  2.   { 
  3.    $project: {
  4.       providedBy: "$medicationDetails.providedBy",
  5.       key: { $cond: [ { $eq: [ "$medicationDetails.subCategory", "" ] }, "$medicationDetails.category", "$medicationDetails.subCategory" ] }
  6.     } 
  7.   },
  8.   {
  9.     $group: { 
  10.       _id: "$key",
  11.       providedByDoctor: { $sum: { $cond: [ { $eq: [ "$providedBy", "Doctor" ] }, 1, 0 ] } },
  12.       providedByNurse: { $sum: { $cond: [ { $eq: [ "$providedBy", "Nurse" ] }, 1, 0 ] } } }
  13.   },
  14.   { $group: { _id: null, medicationDetails: { $push: { name: "$_id", providedByDoctorCount: "$providedByDoctor", providedByNurseCount: "$providedByNurse" } } } },
  15.   { $unset: "_id" }
  16. ])

playground

huangapple
  • 本文由 发表于 2023年6月4日 23:17:23
  • 转载请务必保留本文链接：https://go.coder-hub.com/76401082.html
  • aggregation
  • arrays
  • count
  • group
  • mongodb
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