How to sort nested arrays in JS by number?

Question

I want to sort the nested arrays by salary beginning with the person who receives the highest salary and ending with the person who receives the lowest one. I dont want to take into consideration the currency... I dont know how to do it... Thanks in advance

Array
(
[0]=>Array
(
[firstName]=>John
[lastName]=>Doe
[age]=>35
[salary]=>Array
(
[gbp]=>"180"
[eur]=>“”
[usd]=>""
)
)
[1]=>Array
(
[firstName]=>Maria
[lastName]=>Anders
[age]=>26
[salary]=>Array
(
[gbp]=>""
[eur]=>"100"
[usd]=>""
)
)
[2]=>Array
(
[firstName]=>Thomas
[lastName]=>Hardy
[age]=>31
[salary]=>Array
(
[gbp]=>""
[eur]=>""
[usd]=>"224"
)
)
)

Please, try something and come back after you try. Write what you tried also here so that we can help you. This is not a "do my homework" website. — Israel Zinc
– Israel Zinc, Commented Nov 22, 2017 at 6:30

Get Off My Lawn · Accepted Answer · 2017-11-22 06:49:20Z

1

It looks like you posted php.

Anyways, take a look at the sort function:

var numbers = [{
  age: 10,
  salary: {
    gbp: 10,
    eur: 0,
    usd: 0
  }
}, {
  age: 2,
  salary: {
    gbp: 0,
    eur: 345,
    usd: 0
  }
}, {
  age: 100,
  salary: {
    gbp: 0,
    eur: 0,
    usd: 23
  }
}, {
  age: 50,
  salary: {
    gbp: 0,
    eur: 432,
    usd: 0
  }
}];

numbers.sort(function(a, b) {
  // Get the sum of salaries in the a object
  let aAmt = Object.values(a.salary).reduce((s, v) => s + v, 0)
  // Get the sum of salaries in the b object
  let bAmt = Object.values(b.salary).reduce((s, v) => s + v, 0)
  // Calculate the difference
  return bAmt - aAmt
})

console.log(numbers);

edited Nov 22, 2017 at 6:49

answered Nov 22, 2017 at 6:33

Get Off My Lawn

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Comments

ShdwKnght333 · Accepted Answer · 2017-11-22 06:53:58Z

1

I don't think what you posted is JS, but if you have a similar type of array in JS you could try to use the following

Assuming the array looks something like this

var x = [['John', 'Doe', 35, [180,0,0]],['Maria','Anders',26,[0,100,0]],['Thomas','Hardy',31,[0,0,224]]]

You could then use this

function sorty(a,b){
    if(Math.max.apply(null, a[3]) > Math.max.apply(null, b[3])) return -1;
    if(Math.max.apply(null, a[3]) < Math.max.apply(null, b[3])) return 1;
    return 0
}

Then running this:

x.sort(sorty)

Will give you this:

    0: (4) ["Thomas", "Hardy", 31, Array(3)]
    1: (4) ["John", "Doe", 35, Array(3)]
    2: (4) ["Maria", "Anders", 26, Array(3)]

Sort of a hacky solution I guess.

answered Nov 22, 2017 at 6:53

ShdwKnght333

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Comments

Nina Scholz · Accepted Answer · 2017-11-22 07:44:07Z

While you have only one relevant value in the salary object, you could take logical OR || for getting a value out of it for sorting. If no value is truthy, a zero is taken.

In the callback of Array#sort take the delta of the two values, a - b sorts smaller a before b. If you like to sort descending, just reverse the operands to b - a.

var array = [{ firstName: "John", lastName: "Doe", age: 35, salary: { gbp: "180", eur: "", usd: "" } }, { firstName: "Maria", lastName: "Anders", age: 26, salary: { gbp: "", eur: "100", usd: "" } }, { firstName: "Thomas", lastName: "Hardy", age: 31, salary: { gbp: "", eur: "", usd: "224" } }];

array.sort(function (a, b) {
    var aa = a.salary.gbp || a.salary.eur || a.salary.usd || 0,
        bb = b.salary.gbp || b.salary.eur || b.salary.usd || 0;

    return bb - aa;
});

console.log(array);

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