Reduce multiple ORs in IF statement in JavaScript
Is there a simpler way to rewrite the following condition in JavaScript?
if ((x == 1) || (x == 3) || (x == 4) || (x == 17) || (x == 80)开发者_运维问答) {...}
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.
switch (x) {
case 1:
case 3:
case 4:
case 17:
case 80:
//code
break;
default:
//code
}
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.
a regular expression test uses the string value of x:
if(/^[134]|17|80$/.test(x)){/*...*/}
You can optimize your own example and get rid of a few characters, making it easier on the eyes..:
if (x == 1 || x == 3 || x == 4 || x == 17 || x == 80) { ... }
many options
if ([0, 1, 3, 4, 17, 80].indexOf(x) > 0)
if(/^(1|3|4|17|80)$/.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))...
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
}
Inspired by @Santosh I created a simplified version:
const input = x => [1, 3, 4, 17, 80].includes(x);
console.log(input(10));
console.log(input(1));
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