Is there a simpler way to rewrite the following condition in JavaScript?
if ((x == 1) || (x == 3) || (x == 4) || (x == 17) || (x == 80)) {...}
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8 Answers
You could use an array of valid values and test it with indexOf:
if ([1, 3, 4, 17, 80].indexOf(x) != -1)
Edit Note that indexOf was just added in ECMAScript 5 and thus is not implemented in every browser. But you can use the following code to add it if missing:
if (!Array.prototype.indexOf)
{
Array.prototype.indexOf = function(elt /*, from*/)
{
var len = this.length >>> 0;
var from = Number(arguments[1]) || 0;
from = (from < 0)
? Math.ceil(from)
: Math.floor(from);
if (from < 0)
from += len;
for (; from < len; from++)
{
if (from in this &&
this[from] === elt)
return from;
}
return -1;
};
}
Or, if you’re already using a JavaScript framework, you can also use its implementation of that method.
5 Comments
Daniel Vassallo
+1... This is the neatest. The
in operator does not work correctly for arrays, as you suggested: developer.mozilla.org/en/Core_JavaScript_1.5_Reference/…Daniel Vassallo
@sterofrog: You can add indexOf to older browsers: developer.mozilla.org/en/Core_JavaScript_1.5_Reference/…
user187291
@Daniel: yes, this should be added to the answer, for gooogler's benefit
Joey Adams
Bear in mind that setting Array prototype members breaks the for (i in array) construct, as those prototype members will show up as items of every array. If Array prototype members are set (e.g. with this, by mootools, etc.), you have to iterate over the members with
for var i = 0; i < array.length; i++ , or use a framework function that handles it for you (e.g. jQuery.each()).
Gumbo
@Joey Adams:
for … in is for object properties and not array keys. Always use a the standard ranging for loop for array iteration. (jQuery does that too.)switch (x) {
case 1:
case 3:
case 4:
case 17:
case 80:
//code
break;
default:
//code
}
4 Comments
thecoshman
not exactly simpler, but a nice alternative, and good use of not using a break, which btw you forgot to put after the code in
case 80:
BritishDeveloper
and you need the break in the default case
cmptrgeekken
@BritishDeveloper the default clause doesn't require a break, as there cannot be any cases after the default. For example, see w3schools.com/js/js_switch.asp
BritishDeveloper
@cmptrgeekken true - i had my c# eyes in
This is a little function I found somewhere on the web:
function oc(a) {
var o = {};
for (var i = 0; i < a.length; i++) {
o[a[i]] = '';
}
return o;
}
Used like this:
if (x in oc(1, 3, 4, 17, 80)) {...}
I'm using it for strings myself; haven't tried with numbers, but I guess it would work.
Comments
a regular expression test uses the string value of x:
if(/^[134]|17|80$/.test(x)){/*...*/}
Comments
You can optimize your own example and get rid of a few characters, making it easier on the eyes..:
if (x == 1 || x == 3 || x == 4 || x == 17 || x == 80) { ... }
Comments
many options
if ([0, 1, 3, 4, 17, 80].indexOf(x) > 0)
if(/^(1|3|4|17|80)$/.test(x))
if($.inArray(x, [1, 3, 4, 17, 80])
another one, based on Ed's answer
function list() {
for (var i = 0, o = {}; i < arguments.length; i++)
o[arguments[i]] = '';
return o;
}
if(x in list(1, 3, 4, 17, 80))...
6 Comments
Gumbo
JavaScript’s array index starts with 0.
Gumbo
@stereofrog: Yes:
[0, 1, 3, 4, 17, 80].indexOf(0) === 0.Daniel Vassallo
@stereofrog: so, I think
if ([0, 1, 3, 4, 17, 80].indexOf(x) > 0) needs to be if ([0, 1, 3, 4, 17, 80].indexOf(x) >= 0)user187291
@Daniel Vassallo, this would return 'true' for x=0, which is wrong
Daniel Vassallo
@stereofrog: Ok, I see that you've added an extra element in the array. But why not
if ([1, 3, 4, 17, 80].indexOf(x) != -1)? |
You can also use the Array.includes the simplest way...
if([1,3,4,17,80].includes(x)){
console.log(true);
// rest of the code
}
