Javascript If statement, looking through an array

Question

Mind has gone blank this afternoon and can\'t for the life of me figure out the right way to do this:

if(i!=\"3\" && i!=\"4\" && i!=\"5\" &&a         


        

Answer 1

var a = [3,4,5,6,7,8,9];

if ( a.indexOf( 2 ) == -1 ) { 
   // do stuff
}

indexOf returns -1 if the number is not found. It returns something other than -1 if it is found. Change your logic if you want.

Wrap the numbers in quotes if you need strings ( a = ['1','2'] ). I don't know what you're dealing with so I made them numbers.

IE and other obscure/older browsers will need the indexOf method:

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;  
  };  
}