php - prepend html data in two section of page at once -

so i'll keep simple can bare me, i'm still learning.

i have ajax setup submit code create new category database , prepends page , shows without page refreshing. can pull exact same data on different areas of page @ once.

here stuck. goal finding way prepend 2 different data sets. 1 goes under new categories does, , pulls in field on create new post section of page. thought might element id or can't find on looking for.

so possible take prepend(html) , submit data differently 2 page areas @ once?

here code submission.

<script type="text/javascript" >     $(function() {         $(".update_button").click(function() {             var boxval = $("#category").val();             var boxval = encodeuricomponent(boxval);             var datastring1 = 'category='+ boxval;              var dataconfirm = 'newfaqcategory=true';             var datastring = datastring1+'&'+dataconfirm;             if(boxval=='') {                 alert("please enter title");             } else {                 $.ajax({                     type: "post",                     url: "ajax/faqcn.php",                     data: datastring,                     cache: false,                     success: function(html){                         $("ol#update").prepend(html);                         $("select#faqc_id").prepend(html);                         $("ol#update li:first").slidedown("slow");                         document.getelementbyid('category').value='';                         document.getelementbyid('category').focus();                     }                 });             } return false;         });     });    </script> 

to implement @brett's suggestion, is, instead of sending html code ajax request, send json 2 pieces of html, you'd need this:

php ajax response:

$returnvalue = array(); $returnvalue["html1"] = "<b>html new categories</b>"; $returnvalue["html2"] = "<b>html new post section</b>"; exit(json_encode($returnvalue)); 

then in javascript you'd have:

$.ajax({     type: "post",     url: "ajax/faqcn.php",     data: datastring,     datatype: "json", // type of data you're expecting server     cache: false }) .done(function(data) {     // put html received in right place     $("#html1").prepend(data.html1);     $("#html2").prepend(data.html2); }); 

one last note, "success" callback deprecated of jquery 1.8, it's recommended use .done() instead.