CODEWITHSUNDEEP
phpjqueryajaxmysqli
Patrick
Patrick

Reputation: 11

fetching all rows in my mysql database using ajax and php

Guys i have a problem actually my php code is correct but i dont know why it display undefined into my client side script. Could somebody help me

here is my api.php this is where i put my php

<?php 
 $dbcon = mysqli_connect("localhost", "root", "", "orms") or die("Server not  available" . mysql_error());
 $data = array();
 $result = mysqli_query($dbcon,"SELECT * FROM cottages") or die(mysqli_error());            //query
//$array = mysqli_fetch_row($result);
$data = array();
while($row = mysqli_fetch_assoc($result))
 {
  $data[] = $row;
 }
 echo json_encode( $data )                          //fetch result    
?>

here is my client.php code

   <?php 
     $dbcon = mysqli_connect("localhost", "root", "", "orms") or die("Server not available" . mysql_error());
 ?>
<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 () 
  {
$.ajax({                                      
url: 'api.php', data: "POST", dataType: 'json',  success: function(rows)        
 {
   for (var i in rows)
   {
  var row = rows[i];          

  var cot_id = row[0];
  var image = row[1];
  $('#output').append("<b>cottage: </b>"+cot_id+"<b> image: </b>"+image)
              .append("<hr />");
      } 
     } 
       });

    }); 
  </script>

       </body>
         </html>

Thanks for your help in advance..

Upvotes: 1

Views: 2015

Answers (1)

Barmar
Barmar

Reputation: 780974

You're using mysqli_fetch_assoc, so the row will be an associative array, which turns into a Javascript object. But in the Javascript code you're accessing row as if it's an array, not an object. You need to use the column names:

var cot_id = row.cot_id;
var image = row.image;

(I'm just guessing the column names, because you used SELECT * so I can't see the actual names in your table.)

Upvotes: 1

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