Tips for organizing an Array of Arrays in JSON for improved structure

Question

Tips for organizing an Array of Arrays in JSON for improved structure

Consider the JSON object shown below:

var workers = { "finance" : [   // finance is an array in workers.
                                { "firstName" : "Alice",  // First element
                                  "lastName"  : "Smith",
                                  "age"       : 28 },

                                { "firstName" : "Tom",  // Second Element
                                  "lastName"  : "Brown",
                                  "age"       : 35 }
                              ], // End "finance" array.                                  
              "marketing"       : [ // Marketing is another array in employees.
                                { "firstName" : "Linda", // First Element
                                  "lastName"  : "Taylor",
                                  "age"       : 30 },

                                { "firstName" : "Kevin",   // Second Element
                                  "lastName"  : "Jones",
                                  "age"       : 39 }
                              ] // End "marketing" Array.
            } // End Workers

To access each employee's first name individually, you need to restructure the object as follows:

workers.finance[0].firstName
workers.finance[1].firstName
// etc

javascript arrays json

Answer 1

Answer №1

To restructure this, you would need to remove the "accounting/sales" properties and convert employees into an Array of Objects.

For example: http://jsfiddle.net/hgMXw/

var employees = [
    {
    "dept": "accounting", // new property for this object
    "firstName": "John",
    // First element
    "lastName": "Doe",
    "age": 23
    },
    {
    "dept": "accounting", // new property for this object
    "firstName": "Mary",
    // Second Element
    "lastName": "Smith",
    "age": 32
    },
    {
    "dept": "sales", // new property for this object
    "firstName": "Sally",
    // Third Element
    "lastName": "Green",
    "age": 27
    },
    {
    "dept": "sales", // new property for this object
    "firstName": "Jim",
    // Fourth Element
    "lastName": "Galley",
    "age": 41
    }
]

Answer 2

To restructure this, you would need to remove the "accounting/sales" properties and convert employees into an Array of Objects.

For example: http://jsfiddle.net/hgMXw/

var employees = [
    {
    "dept": "accounting", // new property for this object
    "firstName": "John",
    // First element
    "lastName": "Doe",
    "age": 23
    },
    {
    "dept": "accounting", // new property for this object
    "firstName": "Mary",
    // Second Element
    "lastName": "Smith",
    "age": 32
    },
    {
    "dept": "sales", // new property for this object
    "firstName": "Sally",
    // Third Element
    "lastName": "Green",
    "age": 27
    },
    {
    "dept": "sales", // new property for this object
    "firstName": "Jim",
    // Fourth Element
    "lastName": "Galley",
    "age": 41
    }
]

Answer 3

Answer №2

To handle this situation differently, you need to restructure your code. Consider either incorporating the department as a key in the employee object or accessing it using employees.accounting[0].firstName instead.

If you still want to access the employee as employees[index], you will need to rearrange it as follows:

var employees = [
  { "firstName" : "John", "lastName" : "Doe", "age" : 23, "department" : "accounting" },
  { "firstName" : "...", ..., "department" : "accounting" },
  ... and so forth.
];

Additionally, you can implement a different method for filtering by department.

One approach could be to create a function that iterates through the employees array, filters out matching elements, and returns them in a new array object.

function getAllEmployeesFilteredBy(filterName, filterValue, empArray)
{
  var result = [];
  for (var i=0; i < empArray.length; i++) {
    if (empArray[i][filterName] === filterValue)
      //by reference
      result[result.length] = empArray[i];

      //or by value (creating a new object)
      //result[result.length] = { "firstName" : empArray[i].firstName, "lastName" : empArray[i].lastName, ... }
  }
  return result;
}

Answer 4

To handle this situation differently, you need to restructure your code. Consider either incorporating the department as a key in the employee object or accessing it using employees.accounting[0].firstName instead.

If you still want to access the employee as employees[index], you will need to rearrange it as follows:

var employees = [
  { "firstName" : "John", "lastName" : "Doe", "age" : 23, "department" : "accounting" },
  { "firstName" : "...", ..., "department" : "accounting" },
  ... and so forth.
];

Additionally, you can implement a different method for filtering by department.

One approach could be to create a function that iterates through the employees array, filters out matching elements, and returns them in a new array object.

function getAllEmployeesFilteredBy(filterName, filterValue, empArray)
{
  var result = [];
  for (var i=0; i < empArray.length; i++) {
    if (empArray[i][filterName] === filterValue)
      //by reference
      result[result.length] = empArray[i];

      //or by value (creating a new object)
      //result[result.length] = { "firstName" : empArray[i].firstName, "lastName" : empArray[i].lastName, ... }
  }
  return result;
}

Answer 5

Answer №3

Found on jsFiddle

let staff = { "name" : "Sarah",  // Primary member
                  "position"  : "Manager",
                  "years"       : 10 },

                { "name" : "Michael",  // Supportive member
                  "position"  : "Assistant",
                  "years"       : 5 }
                 ;
alert(staff);

Answer 6

Found on jsFiddle

let staff = { "name" : "Sarah",  // Primary member
                  "position"  : "Manager",
                  "years"       : 10 },

                { "name" : "Michael",  // Supportive member
                  "position"  : "Assistant",
                  "years"       : 5 }
                 ;
alert(staff);

Tips for organizing an Array of Arrays in JSON for improved structure

Answer №1

Answer №2

Answer №3

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