parseInt() behaves differently

Question

I am using the following condition in javascript to filter my options based on postcode.

if (parseInt(postcode, 10) > 99 && parseInt(postcode, 10) < 10000 && postcode != '') {
    // code goes here...
}

where postcode can be any Australian postcode. It works for all except some random postcodes like 0852/0840/4840. I am not sure whether it works for all postcodes I just checked for random postcodes.

Did you try `alert(parseInt(postcode, 10))` for the offending random codes? — Niels Keurentjes, Jul 30 '14 at 07:43
Note: Older browsers will use the octal radix (8) as default when the string begins with "0". As of ECMAScript 5, the default is the decimal radix (10). — trainoasis, Jul 30 '14 at 07:44
parseInt('0852/0840/4840',10) = 852 @OP: Please describe what output do you expect vs what you get — MythThrazz, Jul 30 '14 at 07:45
It's clear from the question that a radix of 10 is already being used. What is not clear is exactly what the value of `postcode` is, what happens and what is expected. — Tom Fenech, Jul 30 '14 at 07:45

score 0 · Answer 1 · answered Jul 30 '14 at 07:53

I just checked and it works absolutely fine for the postcodes you mentioned. Perhaps there are issues in other browsers - that I have not checked anything save or the latest Firefox.

Either way, please make sure the parseInt returns what you expect. You could do something like this:

var parsed = parseInt(postcode, 10);
if (parsed > 99 && parsed < 10000 && postcode != '') {
    // code goes here...
} else {
    console.log(postcode, parsed);
}

The above will print the bad values to the console and you will probably find out what's wrong. Maybe they return NaN? Maybe the input wasn't trimmed?

By the way, I would rework your condition. postcode != '' is a narrow case; what if you input 'abc'? All such values will be NaN after passing them through parseInt, so you could as well do something like this:

var parsed = parseInt(postcode, 10);
if (!!parsed && parsed > 99 && parsed < 10000) {
    // code goes here...
}

Thanks a lot for your help. I figured this out. However the issue was with my google api not with parseint(). Thanks anyways. — user3890489, Jul 31 '14 at 02:26

parseInt() behaves differently

1 Answers1