javascript - I want to keep on storing and displaying the result I am getting from mysql -

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i not expert @ php newbie not sure do.what trying keep on saving result getting textboxthis textbox , shirt,color , price result getting table , keep on displaying on page. picture textbox attached. want keep on storing new result table , keep displaying old one. hope clear trying do. code has 3 files main file products.php has following code

<!doctype html> <html> <head>     <title>products</title>     <p> scan barcode of product </p>     <link href="mystyle.css" rel="stylesheet"> </head> <body>          <input type="text" name="enter" required id="item" style="text-align:centre" placeholder=" barcode id" autofocus />          <div id="item-data"></div>          <script src="js/jquery-2.2.0.min.js"></script> <!--jquery link-->         <script src="js/global.js"></script> <!--linking event file-->  </body> </html>

while code global.js is

$(function(){  //press enter on text area..  $('#item').keypress(function (event) { var key = event.which; if(key == 13)  // enter key code {  var item = $('input#item').val();    // retreiving value item     if ($.trim(item) != ''){            //send php file if     empty or spaces(trim) there dont send it//         $.post('ajax/name.php',{id:item}, function(data){   //using post method sending father(name.php), sending data through file item             $('div#item-data').append(""+data+"</br>"); // grabing data , displaying          });      }  $('#item').val('');  }  });  });

and third file name.php file consists of following code

<?php  require '../db/connect.php';  $id= $_post['id']; // post variable $sql = "select barcodeid, shirts, price clothes  barcodeid=".mysqli_real_escape_string($con,$id);  $result = mysqli_query($con,$sql) or die(mysqli_error($con));  if(mysqli_num_rows($result)>0) {     while($row = $result->fetch_assoc())     {         echo " " . $row["barcodeid"]. " shirt color:  " . $row["shirts"]. "  price: " . $row["price"];     } } else {         echo "id not found, please scan again"; } mysqli_close($con);  ?>

this issue solved using append function in javascript. check in global.js file.


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