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How can I use a variable outside of a click() in jQuery? I have the following code -

$(".area").not(".enemy").click(function(){
                $(".area").removeClass("clicked");
                $(this).toggleClass("clicked");
                attackValue = $(this).text();           
        });

When I want to use attackValue outside of the click() It will not be defined.

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In order to keep attackValue defined outside of the scope of the function, you will need to declare it outside of your click event.

var attackValue;

$(".area").not(".enemy").click(function(){ ... });

Now you should be able to reference it outside.

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If OP declares the attackValue there without the var keyword. It is global and should be accessible outside the function. Although it is impossible to tell using that code snippet.
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For starters: this is NOT how you should declare a variable in javascript:

attackValue = $(this).text();

This would clutter 'programs' variable space. You should 'always' declare a variable in javascript using the var keyword.

When I want to use attackValue outside of the click() It will not be defined.

It depends on where and when you want to use it. Consider the following code:

var attackValue = 'some value';

function setValue() {
  attackvalue = 'some other value';
}

console.log(attackvalue); // this would output some value, because the function hasn;t run yet
setValue();

However if you would do:

var attackValue = 'some value';

function setValue() {
  attackValue = 'some other value';
}

setValue();

console.log(attackValue); // this would output some other value, because the function did already run

Note that the above code should run in a closure. Or it would still clutter the variable space (even when using the var) keyword. So it would look more like:

(function() { // yay closure, now all variables defined in here will stay in here
    var attackValue = 'some value';

    function setValue() {
      attackValue = 'some other value';
    }

    setValue();    

    console.log(attackValue); // this would output some other value, because the function did already run
})();

​

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