Retrieving data from various schemas using Node.js

Question

Retrieving data from various schemas using Node.js

I am managing two schemas, one for employees (parent) and the other for assessments (child). Each assessment is linked to an employee ID with a pass percentage.

Here are some sample data:

employees : [
  {
    "_id": 12345,
    "name": "David",
    "evaluated": false
  },
  {
    "_id": 12346,
    "name": "Miller",
    "evaluated": false
  }
]

Second Schema

assessments: [
  {
    "assessment_type": "basic",
    "employee_id": 12345,
    "qualified": true
  },
  {
    "assessment_type": "advanced",
    "employee_id": 12345,
    "qualified": false
  },
  {
    "assessment_type": "basic",
    "employee_id": 12346,
    "qualified": true
  },
  {
    "assessment_type": "advanced",
    "employee_id": 12346,
    "qualified": true
  }
]

I am trying to retrieve a list of evaluated employees based on their assessments where qualification status is true.

Could you suggest the best approach for achieving this?

node.js mongodb express

Answer 1

Answer №1

Here's an illustration of how we can organize employees based on their assessment results.

const employees = [{
  _id: 12345,
  name: 'David',
  evaluated: false,
}, {
  _id: 12346,
  name: 'Miller',
  evaluated: false,
}];

const assessments = [{
  assessment_type: 'basic',
  employee_id: 12345,
  qualified: true,
}, {
  assessment_type: 'advanced',
  employee_id: 12345,
  qualified: false,
}, {
  assessment_type: 'basic',
  employee_id: 12346,
  qualified: true,
}, {
  assessment_type: 'advanced',
  employee_id: 12346,
  qualified: true,
}];

// Loop through the employees
const sortByAssessment = employees.reduce((tmp, x) => {
  // Retrieve all assessments related to the employee
  const employeeAssessment = assessments.filter(y => y.employee_id === x._id);
  
  // Handle each assessment
  employeeAssessment.forEach((y) => {
    // Process only successful assessments
    if (y.qualified) {
      // Check if this is the first encounter with the assessment_type
      // create an array to store employee information
      tmp[y.assessment_type] = tmp[y.assessment_type] || [];
      
      // Add the employee's name to the assessment type array
      tmp[y.assessment_type].push(x.name);
    }
  });
  
  return tmp;
}, {});

console.log(sortByAssessment);

Answer 2

Here's an illustration of how we can organize employees based on their assessment results.

const employees = [{
  _id: 12345,
  name: 'David',
  evaluated: false,
}, {
  _id: 12346,
  name: 'Miller',
  evaluated: false,
}];

const assessments = [{
  assessment_type: 'basic',
  employee_id: 12345,
  qualified: true,
}, {
  assessment_type: 'advanced',
  employee_id: 12345,
  qualified: false,
}, {
  assessment_type: 'basic',
  employee_id: 12346,
  qualified: true,
}, {
  assessment_type: 'advanced',
  employee_id: 12346,
  qualified: true,
}];

// Loop through the employees
const sortByAssessment = employees.reduce((tmp, x) => {
  // Retrieve all assessments related to the employee
  const employeeAssessment = assessments.filter(y => y.employee_id === x._id);
  
  // Handle each assessment
  employeeAssessment.forEach((y) => {
    // Process only successful assessments
    if (y.qualified) {
      // Check if this is the first encounter with the assessment_type
      // create an array to store employee information
      tmp[y.assessment_type] = tmp[y.assessment_type] || [];
      
      // Add the employee's name to the assessment type array
      tmp[y.assessment_type].push(x.name);
    }
  });
  
  return tmp;
}, {});

console.log(sortByAssessment);

Answer 3

Answer №2

There are two methods to achieve this: you can either join with $look up or populate with the employee id.

assessments.aggregate([
        {
            '$lookup': {
                'from': 'employees',
                'localField': 'employee_id',
                'foreignField': '_id',
                'as': 'datas'
            }
        },
     { "$unwind": "$datas" },


].exec(function(err,result){
   console.log(result)
});

Second approach

// Replace 'assessments' with your model name

      assessments.populate('employee_id').exec(function(err,result){
           console.log(result); 
     });

Answer 4

There are two methods to achieve this: you can either join with $look up or populate with the employee id.

assessments.aggregate([
        {
            '$lookup': {
                'from': 'employees',
                'localField': 'employee_id',
                'foreignField': '_id',
                'as': 'datas'
            }
        },
     { "$unwind": "$datas" },


].exec(function(err,result){
   console.log(result)
});

Second approach

// Replace 'assessments' with your model name

      assessments.populate('employee_id').exec(function(err,result){
           console.log(result); 
     });

Retrieving data from various schemas using Node.js

Answer №1

Answer №2

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