Losing leading 0s when string converts to array

Question

I have a textInput control that sends .txt value to an array collection. The array collection is a collection of US zip codes so I use a regular expression to ensure I only get digits from the textInput.

private function addSingle(stringLoader:ArrayCollection):ArrayCollection {
  arrayString += (txtSingle.text) + '';
  var re:RegExp = /\D/;
  var newArray:Array = arrayString.split(re);

The US zip codes start at 00501. Following the debugger, after the zip is submitted, the variable 'arrayString' is 00501. But once 'newArray' is assigned a vaule, it removes the first two 0s and leaves me with 501. Is this my regular expression doing something I'm not expecting? Could it be the array changing the value? I wrote a regexp test in javascript.

<script type="text/javascript">
  var str="00501"; 
  var patt1=/\D/;
  document.write(str.match(patt1));
</script>

and i get null, which leads me to believe the regexp Im using is fine. In the help docs on the split method, I dont see any reference to leading 0s being a problem.

**I have removed the regular expression from my code completely and the same problem is still happening. Which means it is not the regular expression where the problem is coming from.

FailedDev is correct, use `\d` instead of `\D`. \D means match anything which **isn't** a digit. See the section "Negated Shorthand Character Classes" on this page: http://www.regular-expressions.info/charclass.html — Sly_cardinal, Oct 07 '11 at 00:06
@Shaz arrayString is the String variable I use to store the input and textSingle is the id of the textInput. — smulholland2, Oct 07 '11 at 00:25
@Sly_cardianl I am matching anything that isn't a digit and removing it from the string with the split method. — smulholland2, Oct 07 '11 at 00:27

score 1 · Answer 1 · answered Oct 07 '11 at 00:01

1

What about this ?

/^\d+$/

You can also specify exactly 5 numbers like this :

/^\d{5}$/

answered Oct 07 '11 at 00:01

FailedDev

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stenrap · Answer 2 · 2011-10-07T18:19:47.003

I recommend just getting the zip codes instead of splitting on non-digits (especially if 'arrayString' might have multiple zip codes):

var newArray:Array = [];
var pattern:RegExp = /(\d+)/g;
var zipObject:Object;

while ((zipObject = pattern.exec(arrayString)) != null)
{
    newArray.push(zipObject[1]);
}

for (var i:int = 0; i < newArray.length; i++)
{
    trace("zip code " + i + " is: " + newArray[i]);
}

score 1 · Accepted Answer · answered Oct 07 '11 at 04:59

1

Running this simplified case:

var arrayString:String = '00501';
var re:RegExp = /\D/;
var newArray:Array = arrayString.split(re);
trace(newArray);

Yields '00501' as expected. There's nothing in the code you've posted that would strip leading zeros. You may want to dig around a bit more.

This smells suspiciously like Number coercion: Number('00501') yields 501. Read through the docs for implicit conversions and check if any pop up in your code.

answered Oct 07 '11 at 04:59

hurrymaplelad

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So yeah the code I posted was fine. That array gets passed to another class and that is where the conversion is happening. The docs you recommended helped me find where the coercion was happening. – smulholland2 Oct 07 '11 at 17:28

Losing leading 0s when string converts to array

3 Answers3