Creating a simple Unique Booking ID function

Question

I need a special unique booking ID, which is easy for costumer to read-out over the phone.

It should be 5 characters long, and only use these letters:

var letters = "ABCDEFGHJKMNPQRSTUXY";

I found this function on stackoverflow but is having a hard time to get it to take chars from my letters and limit itself to 5 chars.

var uniqueId = function() {
  return Math.random().toString(36).substr(2, 16);
};

Any help would be appreciated :-)

Why only 5? You're limiting yourself to a small number of "unique" orders. Unless you're expecting to stay small? — evolutionxbox, Jun 15 '17 at 11:27
See [Create GUID / UUID in JavaScript?](https://stackoverflow.com/questions/105034/create-guid-uuid-in-javascript) — Satpal, Jun 15 '17 at 11:28
"small number" ehh.. try and do the math 20x20x20x20x20 :-D if I have that many bookings I promise to expand it to 6 chars :-D — torbenrudgaard, Jun 15 '17 at 11:28
@evolutionxbox no the GUID is completely different from what I am trying to do — torbenrudgaard, Jun 15 '17 at 11:29
Check the duplicate I linked(Can I have my free-stay @Pattaya ? :D ) — Weedoze, Jun 15 '17 at 11:30
`[number of possible char]^[lenght of the generatet id]` - so you'll have 3.200.000 unique ID's... The 3.200.001'th ID won't be unique anymore... Take upper / lower into account if you wasnt to double it... — daan.desmedt, Jun 15 '17 at 11:32
@Weedoze hahaha how did you know?? :-D Great examples except none of them take in account that the same letter must not occur side by side. "ABCAB" is ok - "ABBCD" is not. — torbenrudgaard, Jun 15 '17 at 11:33
@daan.desmedt 3.2 mill bookings would put me in forbes top 400 richest heheh :-D — torbenrudgaard, Jun 15 '17 at 11:36
@torbenrudgaard Edit your post to add this requirement. (I looked at your profile hehe) — Weedoze, Jun 15 '17 at 11:39
@Weedoze if you ever get here write me on thaihome.co.uk and I will hook you up, in trade for some js lessons :-D — torbenrudgaard, Jun 15 '17 at 11:48

Ganhammar · Accepted Answer · 2017-06-15T11:46:42.880

2

See function below, uniqueId() for a string of five characters (it is possible that the function could generate the same string multiple times, so it's not really unique):

function uniqueId(stringLength, possible)
{
  stringLength = stringLength || 5;
  possible = possible || "ABCDEFGHJKMNPQRSTUXY";
  var text = "";

  for(var i = 0; i < stringLength; i++) {
    var character = getCharacter(possible);
    while(text.length > 0 && character === text.substr(-1)) {
      character = getCharacter(possible);
    }
    text += character;
  }

  return text;
}

function getCharacter(possible) {
  return possible.charAt(Math.floor(Math.random() * possible.length));
}

edited Jun 15 '17 at 11:46

answered Jun 15 '17 at 11:37

Ganhammar

1,871
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is there a way to avoid two of the same letters side by side? "ABCAB" is ok - "ABBCD" is not. – torbenrudgaard Jun 15 '17 at 11:38
Yes, see update. – Ganhammar Jun 15 '17 at 11:43
1

Perfect! - exactly what I need - Ganhammar you just made it into our global functions library :-))) thanks. – torbenrudgaard Jun 15 '17 at 11:45
1

@torbenrudgaard Does the loser get a free trip to Pattaya? :p – daan.desmedt Jun 15 '17 at 11:49

score 1 · Answer 2 · answered Jun 15 '17 at 11:35

Something like this

var letters = "ABCDEFGHJKMNPQRSTUXY";
var uniqueId = function() {
  var text = "";
  for (var i = 0; i < 5; i++) {
    text += letters.charAt(Math.floor(Math.random() * letters.length));
  }
  return text;
};
console.log(uniqueId());

[number of possible char]^[lenght of the generatet id] - so you'll have 3.200.000 unique ID's... The 3.200.001'th ID won't be unique anymore...

Take upper / lower cases into account if you want to double the amount of unique ID's...

var letters = "ABCDEFGHJKMNPQRSTUXYabcdefghjkmnpqrstuxy";
    var uniqueId = function() {
      var text = "";
      for (var i = 0; i < 5; i++) {
        text += letters.charAt(Math.floor(Math.random() * letters.length));
      }
      return text;
    };
    console.log(uniqueId());

is there a way to avoid two of the same letters side by side? "ABCAB" is ok - "ABBCD" is not. — torbenrudgaard, Jun 15 '17 at 11:38
Something like this example would solve that - as demo only used AB letters - https://jsfiddle.net/zxos7456/1/ (now taking into account of checking the letter - checking the index could also be a possibility) — daan.desmedt, Jun 15 '17 at 11:41

score 1 · Answer 3 · answered Jun 15 '17 at 11:54

Well some others just beat me, but here's an alternative that is fairly easy to read. It also incorporates not allowing 2 consecutive characters to be the same (as added in a later comment by the OP)

 function uniqueId()
 {
        var letters = "ABCDEFGHJKMNPQRSTUXY";
        var result="";
        while (result.length<5)
        {
            var rand_int = Math.floor((Math.random() * 19) + 1);
            var rand_chr= letters[rand_int];
            if (result.substr(-1, 1)!=rand_chr) result+=rand_chr;
        }
        return (result);
 }

Creating a simple Unique Booking ID function

3 Answers3