ploppy Posted July 25, 2011 Share Posted July 25, 2011 Ths first time i make a post, the post submits fine. I then close the dialog window and click link to reopen and when i submit data again, it duplicates it and adds it twice. This behaviour just seems to keep adding a duplidcte each time. for example, post1 then post1+post1 then post1+post1+post1. Hope you get the idea. If someone could check my code, i would be grateful. Many thanks // Feedback form function feedbacknew() { $("#fb_form").dialog({ autoOpen: false, resizable: true, modal: true, title: 'Submit a feedback request', width: 480, beforeclose: function (event, ui) { $("#fb_message").html(""); }, close: function (event, ui) { $("#fb_message").html(""); $("#feedback").get(0).reset(); } }); $('#fb_submit').click(function () { var name = $('#fb_uname').val(); var client = $('#fb_client').val(); var department = $('#fb_department').val(); var email = $('#fb_email').val(); var position = $('#fb_position').val(); var feedback = $('#fb_feedbacknew').val(); var data = 'fb_uname=' + name + '&fb_client=' + client + '&fb_department=' + department + '&fb_email=' + email + '&fb_position=' + position + '&fb_feedbacknew=' + feedback; $.ajax({ type: "POST", url: "feedback.php", data: data, success: function (data) { $("#feedback").get(0).reset(); $('#fb_message').html(data); //$("#form").dialog('close'); $("#flex1").flexReload(); }, error:function (xhr, ajaxOptions, thrownError){ alert(xhr.status); alert(thrownError); } }); return false; }); $("#fb_form").dialog('open'); } Link to comment https://forums.phpfreaks.com/topic/242733-jquery-dialog-post-being-duplicated/ Share on other sites More sharing options...
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