How to define JavaScript variable inside loop, then call it outside (scope)

Question

I can't seem to get either variable report1 or report2 to have a global scope, I know this problem seems really basic, but I just can't seem to find an answer.

I have to define the variable report1 when numClicks == 1, and then report2 when numClicks == 2. But JavaScript is forgetting the variables as soon as numClicks changes, which is understandable. However, I just cannot seem to widen the scope of the variables. I know for PHP you can do something like return $variable to make it global, but nothing is really working here in JavaScript.

              <?
                "<img  height =\"16\"  src = \"images/stapler.png\"  
                onclick=\"

                    numClicks++;
                if(numClicks == 1){
                    var report1 = '".$reports[$i]."';
                }
                if(numClicks == 2){
                    var report2 = '".$reports[$i]."';
                }
                return confirm(report1);

                \">\n";?>

When the script is executed, on the third click it is returning undefined.

jontsai · Accepted Answer · 2011-08-05 17:55:48Z

3

Whenever you use the var syntax, you are redeclaring the variable.

You need to define the var report1; var report2; outside of your functions in the global scope; within your function, just do a plain assignment (i.e. report2 = ...;)

answered Aug 5, 2011 at 17:55

jontsai

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2 Comments

RD Ward Over a year ago

Doh! I knew it must be simple

Dan Mazzini Over a year ago

Actually, you don't need to declare the variable at all, just use it!

Beska · Accepted Answer · 2011-08-05 17:55:40Z

1

Using "var" will make the variable local to the scope. By eliminating that keyword, it will be global.

(Overly simplified, but that's the basic trick.)

answered Aug 5, 2011 at 17:55

Beska

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Comments

Dan Mazzini · Accepted Answer · 2011-08-05 17:56:31Z

1

To give a global scope to a variable, just omit the "var" keyword:

 report1 = '".$reports[$i]."';

answered Aug 5, 2011 at 17:56

Dan Mazzini

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