bruno_boccara
asked on
Jquery Conflict with prototype
Hello,
I have a problem with a conflict with my script between jquery and prototype.
At the beginning of my script, i use this library
<script src="/SpryAssets/SpryTabbe dPanels.js " type="text/javascript"></s cript>
<script src="/ScriptLibrary/jquery -latest.pa ck.js" type="text/javascript"></s cript>
<script src="/ScriptLibrary/jquery .autocompl ete.js" type="text/javascript"></s cript>
<script src="/ScriptLibrary/jquery .bgiframe. min.js" type="text/javascript"></s cript>
<script src="/lib/prototype.js" type="text/javascript"></s cript>
<script src="/src/scriptaculous.js " type="text/javascript"></s cript>
<script src="/js/ajax.js" type="text/javascript"></s cript>
<script src='/js/jquery/jquery.2.j s' ></script>
I use ajax too :
jQuery.ajax({
url: '/test/test.php',
type: 'POST',
dataType: 'json',
data: 'pRev=' + Rct,
success: function(response)
});
After, i use parseJSON and i have a problem.
The error is : Object function (a,b){return new D.fn.init(a,b)} has no method 'parseJSON'
Here my code : jQuery.parseJSON(response)
Thanks for help.
I have a problem with a conflict with my script between jquery and prototype.
At the beginning of my script, i use this library
<script src="/SpryAssets/SpryTabbe
<script src="/ScriptLibrary/jquery
<script src="/ScriptLibrary/jquery
<script src="/ScriptLibrary/jquery
<script src="/lib/prototype.js" type="text/javascript"></s
<script src="/src/scriptaculous.js
<script src="/js/ajax.js" type="text/javascript"></s
<script src='/js/jquery/jquery.2.j
I use ajax too :
jQuery.ajax({
url: '/test/test.php',
type: 'POST',
dataType: 'json',
data: 'pRev=' + Rct,
success: function(response)
});
After, i use parseJSON and i have a problem.
The error is : Object function (a,b){return new D.fn.init(a,b)} has no method 'parseJSON'
Here my code : jQuery.parseJSON(response)
Thanks for help.
ASKER
I do this but its not work. i do jQuery.noConflict();
but why you do Jquery(document).ready(fun ction($)
{
});
if i do just Jquery. for all its not good ?
but why you do Jquery(document).ready(fun
{
});
if i do just Jquery. for all its not good ?
ASKER CERTIFIED SOLUTION
membership
This solution is only available to members.
To access this solution, you must be a member of Experts Exchange.
ASKER
The problem was because we called twice the jquery library !
Thanks a lot
Thanks a lot
Open in new window
If you need jQuery outside of that then you must use jQuery in place of all $