Sorting array inside array alphabetically

Question

I have an array like this:

const test = [
  {
    label: "C",
    options: [
      { label: "C"},
      { label: "B"}
    ]
  },
  {
    label: "A",
    options: [
      { label: "Z"},
      { label: "Y"}
    ]
  },
  {
    label: "B",
    options: [
      { label: "J"},
      { label: "H"}
    ]
  },
  {
    label: "D",
    options: [
      { label: "T"},
      { label: "B"}
    ]
  }
]

I need to sort alphabetically this array in two ways:

1) By the first label (outside the options array)

2) And the options by label

I only managed to sort by the first label so far, using this example:

function compareStrings(a, b) {
  a = a.toLowerCase();
  b = b.toLowerCase();
  return (a < b) ? -1 : (a > b) ? 1 : 0;
}

test.sort(function (a, b) {
  return compareStrings(a.label, b.label)
})

How can I order the options inside as well?

The expected output array would be:

const test_SORTED = [
  {
    label: "A",
    options: [
      { label: "B"},
      { label: "C"}
    ]
  },
  {
    label: "B",
    options: [
      { label: "Y"},
      { label: "Z"}
    ]
  },
  {
    label: "C",
    options: [
      { label: "H"},
      { label: "J"}
    ]
  },
  {
    label: "D",
    options: [
      { label: "B"},
      { label: "T"}
    ]
  }
]

I got your expected output array. You need to sort the options independent of the parent object — Rajesh Verma
– Rajesh Verma, Commented Mar 19, 2020 at 4:50

Dane Brouwer · Accepted Answer · 2020-03-19 04:52:50Z

3

While sorting the test array, I have sorted the options array also.

const test = [
  {
    label: "C",
    options: [
      { label: "C"},
      { label: "B"}
    ]
  },
  {
    label: "A",
    options: [
      { label: "Z"},
      { label: "Y"}
    ]
  },
  {
    label: "B",
    options: [
      { label: "J"},
      { label: "H"}
    ]
  },
  {
    label: "D",
    options: [
      { label: "T"},
      { label: "B"}
    ]
  }
];
function compareStrings(a, b) {
  a = a.toLowerCase();
  b = b.toLowerCase();
  return (a < b) ? -1 : (a > b) ? 1 : 0;
}

console.log(
    test.sort(function (a, b) {
        a.options = a.options.sort(function (c, d) {
            return compareStrings(c.label, d.label)
        })
        b.options = b.options.sort(function (e, f) {
            return compareStrings(e.label, f.label)
        })
        return compareStrings(a.label, b.label)
    })
)

edited Mar 19, 2020 at 4:52

Dane Brouwer

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answered Mar 19, 2020 at 4:46

Rajesh Verma

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2 Comments

Dane Brouwer Over a year ago

If you're going to use a code snippet, remember to use a console.log to display the output.

Rajesh Verma Over a year ago

Sure, I got it. Thanks @DaneBrouwer

Ilcho Taleski · Accepted Answer · 2020-03-19 04:54:36Z

0

You should sort each of the element's options separately with a forEach loop to avoid sorting them multiple times when doing in sort method.

const comparator = (a, b) => (a.label < b.label ? -1 : a.label > b.label ? 1 : 0);
test.sort(comparator);
test.forEach(el => el.options.sort(comparator));

Also, sort changes the array directly, so no need of assigning values to options.

answered Mar 19, 2020 at 4:54

Ilcho Taleski

1131 silver badge9 bronze badges

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