Reduce multiple ORs in IF statement in JavaScript

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借酒劲吻你
借酒劲吻你 2020-12-07 02:17

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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  • 2020-12-07 02:39

    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) { ... }
    
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  • 2020-12-07 02:40

    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))...
    
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  • 2020-12-07 02:48

    a regular expression test uses the string value of x:

    if(/^[134]|17|80$/.test(x)){/*...*/}
    
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  • 2020-12-07 02:50

    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.

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  • 2020-12-07 02:52

    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.

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  • 2020-12-07 02:53

    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
        }
    
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