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2

In the following JSON object:

var employees = { "accounting" : [   // accounting is an array in employees.
                                { "firstName" : "John",  // First element
                                  "lastName"  : "Doe",
                                  "age"       : 23 },

                                { "firstName" : "Mary",  // Second Element
                                  "lastName"  : "Smith",
                                  "age"       : 32 }
                              ], // End "accounting" array.                                  
              "sales"       : [ // Sales is another array in employees.
                                { "firstName" : "Sally", // First Element
                                  "lastName"  : "Green",
                                  "age"       : 27 },

                                { "firstName" : "Jim",   // Second Element
                                  "lastName"  : "Galley",
                                  "age"       : 41 }
                              ] // End "sales" Array.
            } // End Employees

How do I restructure the object so I can access each employee first name like this:

employees[0].firstName

employees[1].firstName etc.

Thanks!

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3  
This is not a JSON object. It is object literal notation. benalman.com/news/2010/03/theres-no-such-thing-as-a-json . Beside that, do you want to merge sales and accounting into one array? – Felix Kling Jan 19 '11 at 0:53
Yes, I want 1 array that contains several arrays. – Rigil Jan 19 '11 at 0:55
In your example employees[0].firstName, you have one array of objects. Not an array of arrays. Do you mean array of objects? – Felix Kling Jan 19 '11 at 0:56
Yes, an array of objects is what I need. Thanks – Rigil Jan 19 '11 at 0:58

3 Answers

up vote 3 down vote accepted

It would require restructuring it so that you'd eliminate the "accounting/sales" properties and make employees an Array of Objects.

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}
]
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This was what I was looking for. Thanks Patrick – Rigil Jan 19 '11 at 1:01
@Rigil: You're welcome. – user113716 Jan 19 '11 at 1:02
up vote 0 down vote

You can't pivot this like that. Either you move the department as a key in the employee object or you have to access it like employees.accounting[0].firstName.

If you insist on accessing the employee as employees[index], you have to restructure it to:

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

and introduce a different way to filter by department.

maybe create a function that loop through the employees array, and copy each element that match the filter into a new array object and return it.

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

      //or if you prefer by value (different object altogether)
      //result[result.length] = { "firstName" : empArray[i].firstName, "lastName" : empArray[i].lastName, ... }
  }
  return result;
}
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up vote 0 down vote

From jsFiddle

var employees = { "firstName" : "John",  // First element
                  "lastName"  : "Doe",
                  "age"       : 23 },

                { "firstName" : "Mary",  // Second Element
                  "lastName"  : "Smith",
                  "age"       : 32 }
                 ;
alert(employees);
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If you're going to give an answer, please post it here instead of on another site. – user113716 Jan 19 '11 at 1:03
@patrick, @fazo - Alternatively I would post it in both places in case jsfiddle ever goes under. – ChaosPandion Jan 19 '11 at 1:06
@Chaos: Yes, I didn't mean to not include a link at all, but the answer ought to be presented here. The other day, jsFiddle was down for a bit and an asker needed to reference the solutions in a previous question. Unfortunately all the answers given to that question were jsFiddle links! The usefulness of this site shouldn't be subject to the downtime of another. – user113716 Jan 19 '11 at 1:09

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