Run function defined as VAR

Question

I ran into Don't load hidden images ... and wanted to run the function reveal.

$(function () {
       $("img").not(":visible").each(function () {
           $(this).data("src", this.src);
           this.src = "";
       });

       var reveal = function (selector) {
           console.log("Viser billeder");
           var img = $(selector);
           img[0].src = img.data("src");
       }
    });

But how?

If I do this:

reveal();

It returns this in the console:

reveal is not defined

How should I call this?

don't you call it from global context since your variable is of local scope? — heximal, Aug 08 '11 at 19:15
I edited my answer after seeing your jsfiddle. Please take a look :) — Paul, Aug 08 '11 at 19:52

score 4 · Accepted Answer · answered Aug 08 '11 at 19:15

4

Define reveal in the global scope like this, it should work fine.

var reveal;

$(function () {
       $("img").not(":visible").each(function () {
           $(this).data("src", this.src);
           this.src = "";
       });

       reveal = function (selector) {
           console.log("Viser billeder");
           var img = $(selector);
           img[0].src = img.data("src");
       }
    });

answered Aug 08 '11 at 19:15

ShankarSangoli

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It still doesn't work. Calling the `reveal();` still gives me the same error in console. – curly_brackets Aug 08 '11 at 19:27
I want to call the function, when a tab is shown. That's why I need to call it. – curly_brackets Aug 08 '11 at 19:44
There are couple of issue in your fiddle. I fixed it, take a look http://jsfiddle.net/DgL7a/1/ – ShankarSangoli Aug 08 '11 at 19:56
You had defined `reveal` in the outer scope but again specified` var reveal` inside the function. – ShankarSangoli Aug 08 '11 at 19:57
There is still a problem: img[0].src = img.data("src"); - do you know why? – curly_brackets Aug 08 '11 at 20:02
You need to pass some `selector` to reveal function because it take a `selector` param and do `$(selector)`. I have modied the fiddle take a look, hope this makes you understand your mistakes. Take a look http://jsfiddle.net/DgL7a/2/ – ShankarSangoli Aug 08 '11 at 20:07
@Kenneth - Did you get a chance to take a look at this http://jsfiddle.net/DgL7a/2/ ? – ShankarSangoli Aug 08 '11 at 20:33

Paul · Answer 2 · 2011-08-08T19:52:03.113

reveal will only be defined after:

   var reveal = function (selector) {
       console.log("Viser billeder");
       var img = $(selector);
       img[0].src = img.data("src");
   }

And before the }); on the next line, since it is private to the anonymous function.

Edit

After looking at the code from your jsfiddle, you have another error. First of all the reason it's still not working is because after you took @ShankarSangoll's advice and added var reveal in the outer scope, you kept the var keyword in the inner scope as well so now you just have two variables named reveal, the outer one is still undefined. If you remove var from the inner one it will be defined.

However you have a $(function(){ ...}); inside a $(document).ready(function(){ ...}; which is redundant since you know the document is already ready inside your ready function. YOu can change your code to:

$(function() {
    var reveal;
    console.log("Billeder vises");
    $("img").not(":visible").each(function() {
        $(this).data("src", this.src);
        this.src = "";
    });

    var reveal = function(selector) {
        var img = $(selector);
        img[0].src = img.data("src");
    }

    reveal();
});

Or remove the $(function(){ ...}); wrapper altogether since that delays your scripts. You'll also want to pass a selector to reveal(), but I think you know that :)

I need to call the function reveal, when the user clicks on some tabs. How do I call it?? — curly_brackets, Aug 08 '11 at 20:24

score 0 · Answer 3 · answered Aug 08 '11 at 19:18

You are accessing the variable outside of its scope. The var/function (reveal) is defined INSIDE of a parent function ( $(function(){... ), so it can only be used INSIDE of that parent function since it is LOCAL to it (just like any other local variable you define).

Run function defined as VAR

3 Answers3