How to Sort an Array?

Question

If I do this:

console.log(result)

In console I get:

[Object, Object, Object, Object, Object]

Where Object expanded looks like:

contacts: Array[344]
letter: "#"
__proto__: Object

contacts: Array[11]
letter: "A"
__proto__: Object

contacts: Array[31]
letter: "B"
__proto__: Object

contacts: Array[1]
letter: "Z"
__proto__: Object

How can I resort result so that the object with letter == # is at the end of the array?

Thanks

Phrogz · Accepted Answer · 2011-05-02 02:42:27Z

2

result = result.sort(function(a,b){
  a=a.letter;
  b=b.letter;
  return a<b?1:a>b?-1:0;
});

Because '#' is lexicographically before 'a', this sorts by letter reversed, so that your results will come out as "Z, B, A, #". If you want '#' at the end but alphabetics first, perhaps:

result = result.sort(function(a,b){
  a=a.letter;
  b=b.letter;
  var aIsLetter = /[a-z]/i.test(a);
  var bIsLetter = /[a-z]/i.test(b);
  if (aIsLetter && !bIsLetter){
    return -1;
  }else if (bIsLetter && !aIsLetter){
    return 1;
  } else {
    return a<b?-1:a>b?1:0;
  }
});

edited May 2, 2011 at 2:42

answered May 2, 2011 at 2:37

Phrogz

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3 Comments

AnApprentice Over a year ago

Thanks but that ends up making the list start at Z, end in #... I want A through Z and then #

Phrogz Over a year ago

@AnApprentice See the edit. (And in the future, young apprentice, learn to fully specify your needs in your question. :)

AnApprentice Over a year ago

oh glorious that did working, sure takes a lot of code, man oh man. Thanks for your speedy help.

BrunoLM · Accepted Answer · 2011-05-02 02:47:39Z

2

If the letter is lower than A charCode I set it to z char code + 1, see this example on jsFiddle

var list = [{contacts: [], letter: "#", __proto__: {}},
            {contacts: [], letter: "A", __proto__: {}},
            {contacts: [], letter: "Z", __proto__: {}},
            {contacts: [], letter: "B", __proto__: {}}];


list.sort(function(a, b){
    a = a.letter.charCodeAt(0);
    b = b.letter.charCodeAt(0);

    if (a < 65) a = "z".charCodeAt(0) + 1;
    if (b < 65) b = "z".charCodeAt(0) + 1;

    return a>b;
});

$(list).each(function(){    
    $("<span />").html(this.letter).appendTo("pre");
});

answered May 2, 2011 at 2:47

BrunoLM

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