CODEWITHSUNDEEP
javascriptjquery
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Reputation: 2642

Duplicate function calling in javascript

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.

Upvotes: 0

Views: 1571

Answers (2)

aziz punjani
aziz punjani

Reputation: 25776

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

Upvotes: 2

xdazz
xdazz

Reputation: 160833

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

Upvotes: 1

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