Javascript If Statement, Looking Through An Array

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' && i!='6' && i!='

Solution 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;  

    varfrom = Number(arguments[1]) || 0;  
    from = (from < 0)  
         ? Math.ceil(from)  
         : Math.floor(from);  
    if (from < 0)  
      from += len;  

    for (; from < len; from++)  
    {  
      if (frominthis &&  
          this[from] === elt)  
        returnfrom;  
    }  
    return -1;  
  };  
}  

Solution 2:

My mind made this solution:

functionnot(dat, arr) { //"not" functionfor(var i=0;i<arr.length;i++) {
  if(arr[i] == dat){returnfalse;}
}
returntrue;
}

var check = [2,3,4,5,6,7,8,9,18,19,49,50,60,61,70,78,79,80,81,82,90,91,92,93,94]; //numbersif(not(i, check)) {
//do stuff
}

Solution 3:

This solution is cross-browser:

var valid = true;
var cantbe = [3, 4, 5]; // Fill in all your valuesfor (var j in cantbe)
    if (typeof cantbe[j] === "number" && i == cantbe[j]){
        valid = false;
        break;
    }

valid will be true if i isn't a 'bad' value, false otherwise.