MadTechie Posted October 4, 2007 Share Posted October 4, 2007 try this addition $id = (int)$_GET['photo_id']; ///current if(!($_GET['photo_id']))//add { $id = 1; } Quote Link to comment https://forums.phpfreaks.com/topic/71798-solved-question/page/2/#findComment-361754 Share on other sites More sharing options...
adam291086 Posted October 4, 2007 Author Share Posted October 4, 2007 No still get the echoed message. I think i'll just leave it to display the single picture and if they want to view another picture they'll have to click on back. Thanks for the help anyway. Adam Quote Link to comment https://forums.phpfreaks.com/topic/71798-solved-question/page/2/#findComment-361757 Share on other sites More sharing options...
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