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Stop repeating div content in PHP AJAX MySQL

I am trying to retrieve the data dynamically from MySQL database into web page using AJAX technique but the div content is repeated. I want to refresh the div without duplicate the information.

here is the code

api.php

<?php 

require 'db.php';
$query = "SELECT * FROM variables"; 
$result = mysqli_query($con,$query);           
$data = array();
while ( $row = mysqli_fetch_row($result) )
{
  $data[] = $row;
 }
echo json_encode( $data );
?>

client.php

  <html>
   <head>
    <script language="javascript" type="text/javascript" src="jquery.js"> 
     </script>
     </head>
      <body>

        <h2> Client example </h2>
        <h3>Output: </h3>

         <div id="output"></div>

         <script id="source" language="javascript" type="text/javascript">

         function ajaxcall() 
           {
           $.ajax({                                      
            url: 'api.php', data: "", dataType: 'json', timeout:2000, 
           success: function(rows)        
            {
            for (var i in rows)
                {
               var row = rows[i];          
                var id = row[0];
                   var vname = row[1];
              $('#output').append("<b>id: </b>"+id+"<b> name: </b>"+vname)
              .append("<hr />");
               }

              } 
              });
                }; 
                   ajaxcall();
                   setInterval(ajaxcall, (1 * 1000));

                 </script>
                 </body>
                </html>
like image 390
Zahraa Ali Avatar asked Sep 14 '18 11:09

Zahraa Ali


2 Answers

Just change this

 success: function(rows)        
            {
            for (var i in rows)

to this

success: function(rows)        
            {
            $('#output').html(''); // empty the container div and append new HTML
            for (var i in rows)
like image 161
Leonardo Gasparini Avatar answered Nov 06 '22 20:11

Leonardo Gasparini


try to use $('output').empty(); before the loop in order to avoid duplicating content.

like image 23
Ghaith Al-Tameemi Avatar answered Nov 06 '22 19:11

Ghaith Al-Tameemi