I am fairly new to PHP and I am experiencing a problem that I can't find an answer to anywhere!! It really bugs me... I want to post data to a PHP script using AJAX. I am using the JS framework Prototype to do AJAX communication.

İşte JS kodu:

new Ajax.Request("/ondemand/Radio.php", {
          method: 'POST',
          parameters: {programID: id}, 
          onSuccess: function(transport) {
                window.alert("Success, " + id);


          },
          onFailure: function() {
              window.alert("Communication problem");
          },
          onComplete: function() {
              window.alert("Complete");
          }
});

Tüm JS elemanının içinde ... işlevi, bir kutudan bir seçenek seçerken denir

PHP kodu:

<?php 
       //Selects all programs that have a podcast
       $QUERY_SELECT_ALL_PROGRAMS   = "SELECT DISTINCT d.defnr, d.name 
                       FROM definition d, podcast p 
                   WHERE p.program = d.defnr";
       //Select podcasts that belongs to a given program
       $QUERY_SELECT_PODCASTS_FOR_PROGRAM   = "SELECT p.title, p.refnr, p.filename
                           FROM podcast p
                       WHERE program =  ?";
       //Selects all podcasts
       $QUERY_SELECT_ALL_PODCASTS = "SELECT p.refnr, p.title, p.filename, p.filename 
                            FROM podcast";

        $BROADCAST_PATH = "";

        $programID = $_POST["programID"];


/* Returns true if DB connection to server and database is OK
 * Takes mysqli as parameter
 * Connect to the database using the MySQLi API in PHP 5.x
 * This is the prefered way*/
 function DBconnection($connection) {
   $result = false;
   //Refering to $con declared eralier
   //global $connection;
   //Check DB connection
 if ($connection->connect_error) { 
           die('Connect Error: '.$connection- >connect_error); }
else {
    //Refering to $DB_NAME declared earlier
    //Select DB
      global $DB_NAME;
       $DB_selected = $connection->select_db($DB_NAME);
      if (!$DB_selected) { die ('Can\'t use : ' . $connection->connect_error); }
       else { $result = true; }
  }
  return $result;
   }

   ?>


<?php
  echo "<form>";
  echo "<select>";
  //The MySQL connection object, must be created before connection
  $con = new mysqli($MYSQL_SERVER, $MYSQL_USER_NAME, $MYSQL_PASSWORD, $DB_NAME);
if (DBconnection($con)) {
    if ($stmt = $con->prepare($QUERY_SELECT_PODCASTS_FOR_PROGRAM)) {
        $stmt->bind_param("i", $program);
        //$stmt->bind_param("i", $program);
        //$program = $_POST["programs"];
        $program = $_POST["programID"];
        $stmt->execute();
        $stmt->bind_result($title, $refnr, $filepath);
    }
    if (is_null($_POST["programs"])) {
        echo "<option>Choose a program first...</option>";
        //echo "<option>".$file."</option>";
    }
    else {
        if (is_numeric($_POST["programs"])) {
            while($stmt->fetch()) {
                print_r($title);
                //$filepath holds the value of only the name of the broadcast without the entire path
                //40 is the starposition of the name
                $filename = substr($filepath, 40);
                echo "<option value=\"".$refnr."\" id=\"".$refnr."\" onclick=\"play('".$filename."')\">".utf8_encode($title).utf8_encode($filename)."</option>";
            }
        }
    }
     $con->close();
}
echo "</select>";
echo "</form>"; 
  ?>

Here is my problem...The value of $_POST is allways "Array ()". When using a regular form that posts, everything is OK, I get the value, but when using AJAX (not only Prototype), I dont.

Ne i basitçe yapmak istiyorum: AJAX ile Post verileri -> sql sorgusunda kullanımı alınan veri -> sql wuery gelen sonuca dayalı bir HTML lement yapmak ..

Ben POST'ed variabel alamadım bu tür zor

I also took a lokk at what was being sent, and POSTDATA was correct. Please, I really need someones help on this...been looking for days now for an answer..

Daha iyi anlamak için bu yazıyı okuyun .. Same problem