Duplicate function calling in javascript

Question

I have one javascript function that used to display the menu and content in my web page. This is how I call this function :

 $(document).ready(function () {
     ViewProduct(action_name);
 });

So after my page load ready, It will call to this function. But the problem is, I have a search text box used to view the product that the user want to search.

    function SearchClick() { 
        if (typeof select_cat != 'undefined' && typeof select_dep != 'undefined') {
           action_name = "GetProductByCatSearch";
        }else if (typeof select_dep != 'undefined'){
            action_name = "GetProductByDepSearch";
        }else{
            action_name = "GetProductBySearch";
        }
        ViewProduct(action_name);  
    }

So when the SearchClick() is called, it produce the menu and product again. Then there are duplicate menu and content.

what I want is to skip the function in document.ready, when the function in SearchClick() is called.

Welcome to all solutions. Thanks so much.

@MatthewBlancarte : could you give me some example MatthewBlancarte? — Nothing
– Nothing, Commented Mar 29, 2012 at 3:04
@Shyju : It has many line of codes that's why I don't post in here Shyju. — Nothing
– Nothing, Commented Mar 29, 2012 at 3:05

aziz punjani · Accepted Answer · 2012-03-29 02:58:33Z

2

You need to check in viewProduct() if you have already shown the menu for that product then do nothing.

edited Mar 29, 2012 at 2:58

answered Mar 29, 2012 at 2:46

aziz punjani

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Comments

xdazz · Accepted Answer · 2012-03-29 02:42:58Z

1

What you should do is empty the menu first in the ViewProduct function. Function already executed can not be skipped.

answered Mar 29, 2012 at 2:42

xdazz

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