Improving if-else with multiple values?

Question

Is there a better way to do this?

if(cpf.length !== 11 || cpf === "00000000000" || cpf === "11111111111" ||
        cpf === "22222222222" || cpf === "33333333333" || cpf === "44444444444" ||
        cpf === "55555555555" || cpf === "66666666666" || cpf === "77777777777" ||
        cpf === "88888888888" || cpf === "99999999999"){

Technically this is more of a "code review" than an issue.

evolutionxbox
– evolutionxbox

2017-04-18 13:52:27 +00:00
Commented Apr 18, 2017 at 13:52 — evolutionxbox
– evolutionxbox, Commented Apr 18, 2017 at 13:52

Mike Cluck · Accepted Answer · 2017-04-18 13:49:26Z

3

You could debate if this is better but this is what I like to do in that sort of situation:

// Name this something relevant to the problem
var possibleValues = ["0000000000", ...];
if (possibleValues.includes(cpf)) {
  // do stuff
}

or if you're in an environment that doesn't have includes

if (possibleValues.indexOf(cpf) > -1) {
  // do stuff
}

Another possibility is using a regular expression:

if (cpf.length === 11 && cpf.match(/^(\d)\1+$/)) {
  // do stuff
}

^: Start at the beginning
(\d): Look for a digit and remember it
\1+: Look for the remembered digit repeatedly
$: Hit the end of the string

answered Apr 18, 2017 at 13:49

Mike Cluck

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4 Comments

Victor Medeiros Over a year ago

Thanks a lot! Sometimes i get 'tunnel vision' when working improvements for things like this. Also is there any performance gain/loss on the solutions above?. I loved the RegEx solution btw!

Mike Cluck Over a year ago

@VictorMedeiros Glad I could help! Technically, all of these solutions are likely slower than a bunch of ||s. However, it's likely a difference of 1ms or less and unless you really need to squeeze out that last little drop, it's not worth the effort and cruft to optimize it. Always make sure it works and works fast enough before optimizing.

Thomas Over a year ago

to be precise, this part shouldn't be \1{10} rather than \1+, Edit: wait something's wrong here his first condition is cpf.length !== 11 so this whole thing should be if (cpf.length !== 11 || cpf.test(/^(\d)\1{10}$/)) { ... }

Thomas Over a year ago

sry, realized too late that I have a mistake in my edit: it's regex.test(value) or string.match(regex) so ... if (cpf.length !== 11 || cpf.match(/^(\d)\1{10}$/)) { ... }

gurvinder372 · Accepted Answer · 2017-04-18 13:48:08Z

2

Using indexOf Something like

var possibleValues = [ "00000000000", "1111111111" ]; //add more values

if ( cpf.length != 11 || possibleValues.indexOf( cpf ) != -1 )
{
  //value matching
}

answered Apr 18, 2017 at 13:48

gurvinder372

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Comments

RomanPerekhrest · Accepted Answer · 2017-04-18 14:04:01Z

0

Alternative Ecmascript5 solution using isNaN() and RegExp.text() functions:

if (cpf.length !== 11 || (!isNaN(f = cpf[0]) && new RegExp("^"+ f + "{11}$").test(cpf))) {
    // do something
}

isNaN() - to check if we have only numbers(at start)

new RegExp("^"+ f + "{11}$").test(cpf) - to test if we have a sequence of same 11 digits

edited Apr 18, 2017 at 14:04

answered Apr 18, 2017 at 13:58

RomanPerekhrest

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