A Guide to Filtering MongoDB Data Using Array Values

Question

A Guide to Filtering MongoDB Data Using Array Values

I am trying to extract specific data from a document in my collection that contains values stored in an array.

{
  "name": "ABC",
  "details": [
     {"color": "red", "price": 20000}, 
     {"color": "black", "price": 22000}, 
     {"color": "blue", "price": 21000}
  ]
},
{
  "name": "XYZ",
  "details": [
     {"color": "yellow", "price": 10000}, 
     {"color": "black", "price": 12000}, 
     {"color": "green", "price": 11000}
  ]
},
{
  "name": "CBD",
  "details": [
     {"color": "red", "price": 30000}, 
     {"color": "pink", "price": 32000}, 
     {"color": "blue", "price": 31000}
  ]
}

I need help filtering the data for colors red and blue. The desired output is:

{"name": "ABC", "color": "red", "price": 20000},
{"name": "ABC", "color": "blue", "price": 21000},
{"name": "CBD", "color": "red", "price": 30000},
{"name": "CBD", "color": "blue", "price": 31000}

Can someone provide me with the MongoDB query or JS/TS code to achieve this?

javascript angular typescript mongodb mean-stack

Answer 1

Answer №1

If you want to achieve a specific format for your data, you can utilize the flatMap function as shown in the example below:

const information = [
  {
    "name": "XYZ",
    "details": [
       {"color": "green", "price": 25000}, 
       {"color": "yellow", "price": 26000}
    ]
  },
  {
    "name": "DEF",
    "details": [
       {"color": "purple", "price": 35000}, 
       {"color": "orange", "price": 36000}
    ]
  }
];

const modifiedData = information.flatMap(item => item.details.map(detail => ({name: item.name, ...detail})));

console.log(modifiedData);

Answer 2

If you want to achieve a specific format for your data, you can utilize the flatMap function as shown in the example below:

const information = [
  {
    "name": "XYZ",
    "details": [
       {"color": "green", "price": 25000}, 
       {"color": "yellow", "price": 26000}
    ]
  },
  {
    "name": "DEF",
    "details": [
       {"color": "purple", "price": 35000}, 
       {"color": "orange", "price": 36000}
    ]
  }
];

const modifiedData = information.flatMap(item => item.details.map(detail => ({name: item.name, ...detail})));

console.log(modifiedData);

Answer 3

Answer №2

If you want to flatten an array of complex objects with an array attribute, you can achieve this using Array.flatMap(). For reference, you can visit this link.

let data = [
{
  "name": "ABC",
  "details": [
     {"color": "red", "price": 20000}, 
     {"color": "black", "price": 22000}, 
     {"color": "blue", "price": 21000}
  ]
},
{
  "name": "XYZ",
  "details": [
     {"color": "yellow", "price": 10000}, 
     {"color": "black", "price": 12000}, 
     {"color": "green", "price": 11000}
  ]
},
{
  "name": "CBD",
  "details": [
     {"color": "red", "price": 30000}, 
     {"color": "pink", "price": 32000}, 
     {"color": "blue", "price": 31000}
  ]
}
]

data = data.filter(d1 => d1.details.some(d2 => d2.color == "red" || d2.color == "blue"))
data = data.flatMap(({ name,details }) => details.map(d => ({ name, ...d })))

console.log(data)

Answer 4

If you want to flatten an array of complex objects with an array attribute, you can achieve this using Array.flatMap(). For reference, you can visit this link.

let data = [
{
  "name": "ABC",
  "details": [
     {"color": "red", "price": 20000}, 
     {"color": "black", "price": 22000}, 
     {"color": "blue", "price": 21000}
  ]
},
{
  "name": "XYZ",
  "details": [
     {"color": "yellow", "price": 10000}, 
     {"color": "black", "price": 12000}, 
     {"color": "green", "price": 11000}
  ]
},
{
  "name": "CBD",
  "details": [
     {"color": "red", "price": 30000}, 
     {"color": "pink", "price": 32000}, 
     {"color": "blue", "price": 31000}
  ]
}
]

data = data.filter(d1 => d1.details.some(d2 => d2.color == "red" || d2.color == "blue"))
data = data.flatMap(({ name,details }) => details.map(d => ({ name, ...d })))

console.log(data)

Answer 5

Answer №3

If you're looking to achieve optimal performance, this approach could be your go-to solution:

db.collection.aggregate([
   { $match: { "details.color": { $in: ["red", "blue"] } } },
   {
      $set: {
         details: {
            $filter: {
               input: "$details",
               cond: { $in: ["$$this.color", ["red", "blue"]] }
            }
         }
      }
   },
   { $unwind: "$details" },
   { $replaceWith: { $mergeObjects: [{ name: "$name" }, "$details"] } }
])

Explore Mongo Playground

Answer 6

If you're looking to achieve optimal performance, this approach could be your go-to solution:

db.collection.aggregate([
   { $match: { "details.color": { $in: ["red", "blue"] } } },
   {
      $set: {
         details: {
            $filter: {
               input: "$details",
               cond: { $in: ["$$this.color", ["red", "blue"]] }
            }
         }
      }
   },
   { $unwind: "$details" },
   { $replaceWith: { $mergeObjects: [{ name: "$name" }, "$details"] } }
])

Explore Mongo Playground

Answer 7

Answer №4

If you're not concerned about performance, you have the option to execute the following JavaScript code:

let dataArr = [{
    "name": "XYZ",
    "details": [
       {"color": "green", "price": 50000}, 
       {"color": "yellow", "price": 51000}
    ]
  },

  {
    "name": "EFG",
    "details": [
       {"color": "purple", "price": 40000}, 
       {"color": "orange", "price": 41000}
    ]
  }
]
let finalArr = []
for (let j = 0; j < dataArr.length; j++) {
    for (let m = 0; m < dataArr[j].details.length; m++) {
        finalArr.push({title: dataArr[j].name, ...dataArr[j].details[m]})
    }
}

Answer 8

If you're not concerned about performance, you have the option to execute the following JavaScript code:

let dataArr = [{
    "name": "XYZ",
    "details": [
       {"color": "green", "price": 50000}, 
       {"color": "yellow", "price": 51000}
    ]
  },

  {
    "name": "EFG",
    "details": [
       {"color": "purple", "price": 40000}, 
       {"color": "orange", "price": 41000}
    ]
  }
]
let finalArr = []
for (let j = 0; j < dataArr.length; j++) {
    for (let m = 0; m < dataArr[j].details.length; m++) {
        finalArr.push({title: dataArr[j].name, ...dataArr[j].details[m]})
    }
}

Answer 9

Answer №5

Utilizing the power of aggregation, this query can be incredibly beneficial.

db.collection.aggregate([
  {
    $unwind: "$details"
  },
  {
    $match: {
      "details.color": { $in: ["red", "blue"] }
    }
  },
  {
    $project: {
      _id: 0,
      name: 1,
      color: "$details.color",
      price: "$details.price"
    }
  }
]);

Answer 10

Utilizing the power of aggregation, this query can be incredibly beneficial.

db.collection.aggregate([
  {
    $unwind: "$details"
  },
  {
    $match: {
      "details.color": { $in: ["red", "blue"] }
    }
  },
  {
    $project: {
      _id: 0,
      name: 1,
      color: "$details.color",
      price: "$details.price"
    }
  }
]);

Answer 11

Answer №6

Utilizing MongoDB's $unwind operator can be beneficial for -

Breaking down an array field within the input documents to generate a document for each element. Each resulting document consists of the input document with the value of the array field substituted by the specific element.

An example of an aggregate query is shown below -

db.getCollection('your_collection').aggregate([
  // create a document for each element in `details` array
  { $unwind : "$details" },

  // filter document with color red or blue
  { $match: { "details.color": { $in: ["red", "blue"] } } },
  
  // output in desired format
  { $project: { name: 1, color: "$details.color", price: "$details.price" } }
])

This showcases how the deconstructed data might appear -

[
  {"name": "ABC", "details": { "color": "red", "price": 20000} },
  {"name": "ABC", "details": { "color": "blue", "price": 22000} },
  {"name": "ABC", "details": { "color": "blue", "price": 21000} },
  {"name": "XYZ", "details": { "color": "yellow", "price": 10000} },
  {"name": "XYZ", "details": { "color": "black", "price": 12000} },
  ...
]

Filtered and deconstructed data (limited to red & blue) -

[
  {"name": "ABC", "color": "red", "price": 20000},
  {"name": "ABC", "color": "blue", "price": 21000},
  {"name": "CBD", "color": "red", "price": 30000},
  {"name": "CBD", "color": "blue", "price": 31000}
]

Answer 12

Utilizing MongoDB's $unwind operator can be beneficial for -

Breaking down an array field within the input documents to generate a document for each element. Each resulting document consists of the input document with the value of the array field substituted by the specific element.

An example of an aggregate query is shown below -

db.getCollection('your_collection').aggregate([
  // create a document for each element in `details` array
  { $unwind : "$details" },

  // filter document with color red or blue
  { $match: { "details.color": { $in: ["red", "blue"] } } },
  
  // output in desired format
  { $project: { name: 1, color: "$details.color", price: "$details.price" } }
])

This showcases how the deconstructed data might appear -

[
  {"name": "ABC", "details": { "color": "red", "price": 20000} },
  {"name": "ABC", "details": { "color": "blue", "price": 22000} },
  {"name": "ABC", "details": { "color": "blue", "price": 21000} },
  {"name": "XYZ", "details": { "color": "yellow", "price": 10000} },
  {"name": "XYZ", "details": { "color": "black", "price": 12000} },
  ...
]

Filtered and deconstructed data (limited to red & blue) -

[
  {"name": "ABC", "color": "red", "price": 20000},
  {"name": "ABC", "color": "blue", "price": 21000},
  {"name": "CBD", "color": "red", "price": 30000},
  {"name": "CBD", "color": "blue", "price": 31000}
]

A Guide to Filtering MongoDB Data Using Array Values

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

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