Switch Statement in Javascript

Question

Can anyone help me?

This doesn't seem to work:

switch (parseInt(charCode))
    {
        case (charCode >= 65 && charCode <=90): //UPPERCASE
            alert("UP");
        break;

        case (charCode >= 97 && charCode <=122): //LOWERCASE
            alert("LO");
        break;

        case (charCode >= 48 && charCode <=57): //NNUMBERS
            alert("NUM");
        break


    }

Thanks

What does it currently do? What do you want to achieve? What's in parseInt(charCode)? — Ondřej Mirtes
– Ondřej Mirtes, Commented Jun 18, 2011 at 9:49
Switch can only be used to check against specific values, not ranges. Read this for more info: developer.mozilla.org/en/JavaScript/Reference/Statements/switch — tcooc
– tcooc, Commented Jun 18, 2011 at 9:57

Community · Accepted Answer · 2017-05-23 10:08:06Z

5

Cases must be single values, not expressions. If you need to use expressions, use an if else if series:

var numCharCode = parseInt(charCode, 10); // Always best to include the radix argument
if (numCharCode >= 65 && numCharCode <=90) {       //UPPERCASE
    alert("UP");
}
else if (numCharCode >= 97 && numCharCode <=122) { //LOWERCASE
    alert("LO");
}
else if (numCharCode >= 48 && numCharCode <=57) {  //NNUMBERS
    alert("NUM");
}

(Always assuming charCode really is a string in the first place. The name suggests it's already a number.)

Off-topic: Your logic is English-centric (and fails even for English, on words we've copied from other languages, like "naïve"). If that's okay, great, but if not you'll want to look more into how you reliably determine the case of a character in a language-neutral way. More in this answer here on StackOverflow.

edited May 23, 2017 at 10:08

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answered Jun 18, 2011 at 9:48

T.J. Crowder

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3 Comments

thebounder Over a year ago

Have remembered how to do it myself: charCode = parseInt(charCode); switch (true) { case (charCode >= 65 && charCode <=90): //UPPERCASE // return true; alert("UP"); break; case (charCode >= 97 && charCode <=122): //UPPERCASE // return true; alert("LO"); break; case (charCode >= 48 && charCode <=57): //NUMBERS // return true; alert("NUM"); break; case (charCode == 8 || charCode == 13): return true; break; default: return false; break; }

Lightness Races in Orbit Over a year ago

Actually it even fails for some "original" English words (definition tricky due to how the language came to be). "encyclopædia" comes to mind.

T.J. Crowder Over a year ago

@Tomalek: "encyclopædia" is copied from Greek, indirectly via Latin, which is why it has the ash diphthong ("æ") which is fairly common in transliterations of words from the Greek to Latin alphabets. (English is such a mutt. :-) ) But again, exactly, this is an English word and has been for nearly 500 years, and for most of that time it was spelt with the diphthong.

KooiInc · Accepted Answer · 2011-06-18 16:02:39Z

1

[edit] Let's add this disclaimer: this use of switch is considered EVIL or ABUSIVE. Some Disciples Of The Javascript also consider the use of ternaries as a major SIN. So beware of using them, because hell may await you, so the Gospel goes.

You can use switch like this ¹:

switch (true) {
   case (parseInt(charCode) >= 65 && parseInt(charCode) <=90): //UPPERCASE
       alert("UP");
       break;
   case (parseInt(charCode) >= 97 && parseInt(charCode) <=122): //LOWERCASE
       alert("LO");
       break;
   case (parseInt(charCode) >= 48 && parseInt(charCode) <=57): //NNUMBERS
       alert("NUM");
       break;
   default: break
}

it's pretty bulky. You don't have to use parseInt if you derive charCode came from event.keyCode. If you need to use parseInt, don't forget to provide the radix, or better still, use Number to convert to a number.

Anyway using a ternary is an alternative for this:

alert( charCode >= 65 && charCode <= 90 
        ? 'UP'
        : charCode >= 97 && charCode <= 122
        ? 'LO'
        : charCode >= 48 && charCode <= 57 
        ? 'NUM'
        : 'OTHER'
);

[edit] Let's see if the following alternative can satisfy the church of Javascript Coders...

Another alternative is using a RegExp with the character derived from the keyCode:

var chr      = String.fromCharCode(charCode),
    alertval =    /[a-z]/.test(chr) ?  'LO'
                : /[A-Z]/.test(chr) ?  'UP'
                : /[0-9]/.test(chr) ?  'NUM'
                : 'OTHER';
    alert(alertval);

As a last alternative (man, javascript is so versatile!) I present:

var check = String.fromCharCode(charCode)
             .replace(/([a-z])|([A-Z])|([0-9])|(.+)/,
                function(a,b,c,d) {
                  return b ? 'LO' : c ? 'UP' : d ? 'NUM' : 'OTH';
             });

¹ some notes on that 'trick'

edited Jun 18, 2011 at 16:02

answered Jun 18, 2011 at 9:54

KooiInc

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12 Comments

T.J. Crowder Over a year ago

What ... a ... horrible ...idea. :-) Still, though, it should work, cases are evaluated in order.

thebounder Over a year ago

Thanks, I remembered that's how to do it. It is the exact correct method and far better than a string of ugly if else statements.

T.J. Crowder Over a year ago

@thebounder: Ugliness, like beauty, is in the eye of the beholder. :-)

mplungjan Over a year ago

I remember now that I forgot that and I am so happy I did forget it.

KooiInc Over a year ago

@all: that's why I provided an alternative ;~)

|

Dmitri Sologoubenko · Accepted Answer · 2011-06-18 10:07:30Z

1

You shouldn't take as true the supposition about Javascript uses ASCII character table (doesn't it use UTF8?), so instead of numerical value comparison, just use character (string) value comparison. You shouldn't use switch statement in such a way: as in your example, it performs a (value and type, "===") comparison between parseInt(charCode) (which should be an integer) and boolean expression values such that returned by your case statements. Use and IF chain, as follows (supposing charCode is still a string value):

if (charCode >= 'A' && charCode <='Z') { //UPPERCASE
            alert("UP");
} else if (charCode >= 'a' && charCode <='z') { //LOWERCASE
            alert("LO");
        break;
} else if (charCode >= '0' && charCode <='9'): //NNUMBERS
            alert("NUM");
}

For ASCII numerical value of a char, use Javascript "asc" function, and so if you charcode is coming from a key event, you can use the inverse, the chr(keyCode) function in order to obtain a string (character) value of your already numerical keycode. I hope it was helpful :)

edited Jun 18, 2011 at 10:07

answered Jun 18, 2011 at 10:01

Dmitri Sologoubenko

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4 Comments

mplungjan Over a year ago

I think you need to do String.fromCharCode here

T.J. Crowder Over a year ago

"You shouldn't take as true the supposition about Javascript uses ASCII character table (doesn't it use UTF8?)" It uses UTF-16, actually. But it doesn't matter in this case. ASCII and Unicode share the same first 127 values, by design. The bigger issue is that this is a very naïve definition of "upper case" and "lower case" that fails easily. For instance, looking at that last sentence, is the "ï" in "naïve" lower case? I think it is, I'm pretty sure the French think it is, but this code does not. (And "naïve" is an English word, we just copied it from the French a while back.)

Lightness Races in Orbit Over a year ago

@TjCrowder: And the non-borrowed "encyclopædia" vs "ENCYCLOPÆDIA".

Yevgeny Simkin Over a year ago

how does one type Æ on a keyboard?(I cut and paste it)

mplungjan · Accepted Answer · 2011-06-18 13:00:45Z

0

don't use switch here - it is not correct syntax and charCode is an int so no need to parseInt it

    if (charCode >= 65 && charCode <=90) //UPPERCASE
        alert("UP");
    else if (charCode >= 97 && charCode <=122) //LOWERCASE
        alert("LO");
    else if (charCode >= 48 && charCode <=57) // NUMBERS
        alert("NUM");

UPDATE: How about fromCharCode

var char = String.fromCharCode(charCode)

if (/[a-z]/.test(char)) alert("Lower");
else if (/[A-Z]/.test(char)) alert("Upper");
else if (/[0-9]/.test(char)) alert("Numbers");

http://jsfiddle.net/mplungjan/v65Xc/

I am sure someone can put this in an object or use prototype :)

edited Jun 18, 2011 at 13:00

answered Jun 18, 2011 at 9:48

mplungjan

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5 Comments

thebounder Over a year ago

Hi, charCode is an int, and Swicth is the exact syntax to use rather than an ugly string of if else's.

T.J. Crowder Over a year ago

@thebounder: If charCode is an int, why are you calling parseInt on it?

thebounder Over a year ago

There are only three in the example I gave.

KooiInc Over a year ago

Help!

if (/[a-z]/.test(char) alert("Lower"); else if (/[A-Z]/.test(char) alert("Upper"); else if (/[0-9]/.test(char) alert("Numbers");

doesn't pass JsHint.com

mplungjan Over a year ago

missing end brackets. Thanks. Fixed

Dmitri Sologoubenko · Accepted Answer · 2011-06-18 14:15:02Z

0

Please visit this website for functions and explanation related to your problem. Sorry but none of them takes in consideration the problem related to non-ascii chars like 'ï' or 'ö' http://javascript.about.com/library/blvalid02.htm

answered Jun 18, 2011 at 14:15

Dmitri Sologoubenko

2,9191 gold badge24 silver badges28 bronze badges

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