php - Show div on successful response -


i'm trying success div show after php data sent server, updating div isn't showing.

can show me i'm going wrong. i've looked around stack overflow , tried many different solutions no luck. 

<div id= "infomsg" style="display: none;">updated successfully! </div>  $.ajax({ type: "post", url: "validate.php", cache: false, datatype: "json", data: datastring, success: function(success){   $('#infomsg').show(success)         }         });   if (isset($_post)) {       $first=$_post['first'];     $middle=$_post['middle'];     $last=$_post['last'];     $email=$_post['email'];     $lives=$_post['lives'];     $gender=$_post['gender'];     $about=$_post['about'];     $fav_track=$_post['fav_track'];     $fav_bands=$_post['fav_bands'];   $sql3 = "update user set first='$first',middle='$middle',last='$last',email='$email',lives='$lives',gender='$gender', about='$about', fav_track='$fav_track', fav_bands='$fav_bands' `id`='".$_session['id']."'";         $res3 = mysqli_query($mysqli,$sql3) or die(mysqli_error($mysqli));           //writes photo server   if($res3)   {   //tells if ok   echo '<div id= "infomsg" style="display: none;">updated successfully! </div>';   }else{    //gives , error if not   echo "sorry, there problem updating information, contact se7ern@se7ern.co.uk ref:#1112-validate.";   }      }

if want show <div> not have pass parameter in "show()" method.

you can below mentioned approach :

<div id="infomsg" style="display: none;">updated successfully!</div>  $.ajax({     type: "post",     url: "validate.php",     cache: false,     datatype: "json",     data: datastring,     success: function (success) {         $('#infomsg').show();     } });

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