Question

I have some jquery ajax code to update a record but it’s not working as expected.

Here is the code:

    function update_records() {
    $.ajax({
  type: "POST", // Set the type then $_GET['id'] or $_POST['id']
  url: "update_record.php",
  data: { category: "John", image: "Boston" }
}).done(function( msg ) {
  alert( "Data Saved: " + msg );
});

}; //end function


Then the php

<?php 

  $id = $_REQUEST['id'];


  include 'config.php';


  $result = mysql_query("UPDATE mytable SET title = 'something' where id=$id"); 
  $result = mysql_query("SELECT * FROM mytable WHERE id = '$id'");             
  $array = mysql_fetch_row($result);  

?>

Answer 1

Example from jQuery.com
http://api.jquery.com/jQuery.ajax/

$.ajax({
  type: "POST", // Set the type then $_GET['id'] or $_POST['id']
  url: "some.php",
  data: { name: "John", location: "Boston" }
}).done(function( msg ) {
  alert( "Data Saved: " + msg );
});

I can’t understand what your query returns.
Maybe you need to select where id = $id;

 //  $id = $_POST['id'] or $_GET['id']  Where is $id??? 
  $result = mysql_query("UPDATE mytable SET title = 'something' where id=$id"); 
  $result = mysql_query("SELECT * FROM mytable WHERE id = '$id'");             
  $array = mysql_fetch_row($result);    


 //You can use mysql_error() function to see the error.
 $result = mysql_query("UPDATE mytable SET title = 'something' where id=$id") or    die(mysql_error());