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How can I rewrite my code below if my 'data_select.php' is in a variable $data = dataselect, that when it is set to true the data_select.php will show. I'm making a dynamic content in php and I'm not familiar with jquery. Any idea? thank you in advance. <script type="text/javascript"> $(document).ready(function() { $('a').click(function(){ $.get('data_select.php', {'page':$(this).text()}, function(data){ $('#content').html(data); }); }); }); </script> When I try firebug, the error says 'data_select.php' not found.
