CODEWITHSUNDEEP
javascriptjqueryfirebug
Imran Subhani
Imran Subhani

Reputation: 1104

I have function but in firebug console it says its not a function where I am calling it?

I am using a function in two sites but when I want to implement it to third site it does not work. When I look into firebug console it says its not a function. My function is in a separate file called profilter.js look like this:

jQuery.fn.sfProductFilter = function (options) {
    options = options || {};
    return this.each(function () {
        var pf = new SFProductFilter(this, options)
    })
}

and I am calling it from a page and code is: 

$(document).ready(function(){
            console.log($("ul.productSmall"));
            $('ul.productSmall').sfProductFilter(); //says not a function.
});

I have checked through console.log followings 1- jQuery is included already 2- If I console.log from the js file it works but inside any code block it does not 3- ul.productSmall shows right results in console.log

I can provide link of site but just not providing it so moderator wont thinks its a spam.

I have struggled a lot please let me know where I am making mistake?

console look like

Upvotes: 0

Views: 597

Answers (2)

user1233508
user1233508

Reputation:

(Copied from comment above.)

You are including jQuery twice, the second load happens far below the inclusion of profilter.js and destroys your custom function.

Upvotes: 2

mc.
mc.

Reputation: 549

Are you including the jquery ui script? Looks like .button() is called in 

function myChecks(){
    $("#checkboxcontainer input[type='checkbox']").button();
}

I have a hunch this will do the trick.

Upvotes: 0

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