javascript - jQuery Ajax data to same PHP page not working as INTENDED? -

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i have 22.php both form , php script reside.

problems;

before submit screenshot

after submit screenshot

1) when submit, result shows below. duplicates form.

2) further entering in top (above) form, changes result accordingly.

3) when enter in bottom form, result changes accordingly , disappear bottom from.

what have tried far solutions;

1) removed totally - url: '',

2) replaced same page - url: '22.php',

3) replaced - var yourdata = $(this).serialize();

4) placed php script after body tag

none of above solve! please help!

<html>          <head>        <title>my first php page</title>        <script type="text/javascript"            src="http://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>        <script>        $(document).ready(function() {                      $("#btn").click(function(event){             event.preventdefault();              		    var myname = $("#name").val();        		var myage = $("#age").val();              var yourdata ='name='+myname+'&age='+myage; // php expecting name , age              $.ajax({                type:'post',  			  data:yourdata,//without serialized                                //url: '22.php',                success:function(data) {                   if(data){                      $('#testform')[0].reset();//reset form          			$('#result').html(data); // here html()                      //alert('submitted');                   }else{                      return false;                    }                                         }               });  		                         });                        });      </script>     </head>      <body>    <?php  if ( isset($_post['name']) ) { // form submitted?      echo "welcome ". $_post["name"] . "<br>";      echo "you ". $_post["age"] . "years old<br>";  }  ?>  	        <form method="post" id="testform">          name:          <input type="text" name="name" id="name" />age:          <input type="text" name="age" id="age" />          <input type="submit" name="submit" id="btn" />        </form>  	<div id='result'></div>  	        </body>  </html>

just place php code top , put exit();. here full code:

<?php if ( isset($_post['name']) ) { // form submitted?     echo "welcome ". $_post["name"] . "<br>";     echo "you ". $_post["age"] . "years old<br>";     exit; } ?> <html>         <head>       <title>my first php page</title>       <script type="text/javascript"           src="http://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>       <script>       $(document).ready(function() {                     $("#btn").click(function(event){            event.preventdefault();              var myname = $("#name").val();             var myage = $("#age").val();             var yourdata ='name='+myname+'&age='+myage; // php expecting name , age             $.ajax({               type:'post',               data:yourdata,//without serialized                               //url: '22.php',               success:function(data) {                  if(data){                     $('#testform')[0].reset();//reset form                     $('#result').html(data); // here html()                     //alert('submitted');                  }else{                     return false;                   }                                        }              });              });                       });     </script>    </head>     <body>    <form method="post" id="testform">     name:     <input type="text" name="name" id="name" />age:     <input type="text" name="age" id="age" />     <input type="submit" name="submit" id="btn" />   </form>   <div id='result'></div>     </body> </html>

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