使用ajax提交表单并发布到页面

Question

A bit of a tricky one here Is there a way i can join the two scripts below into one so when the form is posted it actually loads both the different named url's into the 2 different named divs

    <script type="text/javascript">
$(document).ready(function(){
$("#profile").validate({
     debug: false,
    submitHandler: function(form) {$.post('brides_Includes/edit-profile-top.php', $("#profile").serialize(), function(data) {
       $('#results').html(data);
        //This executes the JavaScript passed back by the ajax.
        $("#results").find("script").each(function(i) {
          eval($(this).text());
        });

        $("form#profile")[0].reset();
      });
    }
  });
});
</script>

<script type="text/javascript">
$(document).ready(function(){
$("#profile").validate({
     debug: false,
    submitHandler: function(form) {$.post('brides_Includes/welcome-menu.php', $("#profile").serialize(), function(data) {
       $('#mymenu').html(data);
        //This executes the JavaScript passed back by the ajax.
        $("#mymenu").find("script").each(function(i) {
          eval($(this).text());
        });

        $("form#profile")[0].reset();
      });
    }
  });
});
</script>

I am not sure if this is even possible but would be great if it was :) Many thanks

Answer 1

Just put the contents of the second submitHandler into the first one:

submitHandler: function(form) {
   var jqxhr1 = $.post('brides_Includes/edit-profile-top.php', $("#prof...
   function(data) {
       $('#results').html(data);
        //This executes the JavaScript passed back by the ajax.
        $("#results").find("script").each(function(i) {
          eval($(this).text());
        });
      });
    }
  var jqxhr2 = $.post('brides_Includes/welcome-menu.php', $("#pro...
  function(data) {
   $('#mymenu').html(data);
    //This executes the JavaScript passed back by the ajax.
    $("#mymenu").find("script").each(function(i) {
      eval($(this).text());
    });

  $.when(jqxhr1, jqhxr2).done(function() { $("#profile")[0].reset(); });
}