Aggiungere Rimuovere Favoriti

[blackskyisback]

Salve a tutti

ho trovato uno script in rete per aggiungere e rimuovere una icona per preferiti in Ajax

il tutto e' stato adattato al mio codice e al database, ma per qualche strano errore (che non riesco a trovare non visualizzo l'icona per aggiungere e rimuovere)



partiamo dal database, la tabeella favs e' collegata alla tabella actor e al campo actor_id

CREATE TABLE `favs` (
  `favoriti_id` smallint(5) unsigned NOT NULL,
  PRIMARY KEY (`favoriti_id`),
  CONSTRAINT `favs_ibfk` FOREIGN KEY (`favoriti_id`) REFERENCES `actor` (`actor_id`) ON DELETE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;

codice index.php

 <script src="jquery.min.js"></script>
    <script type="text/javascript">
    jQuery(document).ready(function($){
          $('.button').on('click', function(e){
              e.preventDefault();
              var actor_id = $(this).attr('actor_id'); // Get the parameter user_id from the button
              var favoriti_id = $(this).attr('favoriti_id'); // Get the parameter director_id from the button
              var method = $(this).attr('method');  // Get the parameter method from the button
              if (method == "Like") {
                $(this).attr('method', 'Unlike') // Change the div method attribute to Unlike
                $('#' + favoriti_id).replaceWith('<img class="favicon" id="' + favoriti_id + '" src="favon.jpg">') // Replace the image with the liked button
              } else {
               $(this).attr('method', 'Like')
               $('#' + favoriti_id).replaceWith('<img class="favicon" id="' + favoriti_id + '" src="favoff.png">')
              }
              $.ajax({
                  url: 'favs.php', // Call favs.php to update the database
                  type: 'GET',
                  data: {actor_id: actor_id, favoriti_id: favoriti_id, method: method},
                  cache: false,
                  success: function(data){
                  }
              });
          });
      });
    </script>



<?php
      $servername = "localhost";
      $username = "xxx";
      $password = "xxx";
      $dbname = "xxx";
      // Create connection
      $conn = new mysqli($servername, $username, $password, $dbname);
      // Check connection
      if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
      }

      function checkFavorite($actor_id, $favoriti_id, $conn) {
        $result = $conn->query("SELECT * FROM favs WHERE actor_id = '". $actor_id."' AND favoriti_id = '". $favoriti_id."'");
        $numrows =  $result->num_rows;
        if ($numrows == 0) {
         echo "<div class = 'button' method = 'Like'  actor_id = ".$actor_id." favoriti_id = ".$favoriti_id."> <img id=".$favoriti_id." src='favoff.png'> </div>";
        }
        else {
          echo  "<div class = 'button' method = 'Unlike'  actor_id = ".$actor_id." favoriti_id = ".$favoriti_id."> <img id=".$favoriti_id." src='favon.jpg'> </div>";
        }
      }
      // Query to get the user_id
      $result = $conn->query("SELECT * FROM actor WHERE nome = 'Alex'");
      $row = $result->fetch_assoc();
      $actor_id = $row['actor_id'];

      // Query to Get the Director ID
      $result = $conn->query("SELECT * FROM actor WHERE nome = 'Alex'");
      $row = $result->fetch_assoc();
      $actor_id = $row['actor_id'];

      echo "<p>Actor: ".$row['nome']."</p> ";
      $fav_image = checkFavorite($actor_id, $favoriti_id, $conn);
      echo "Favorite? : ".$fav_image."";
    ?>

secondo codice favs.php per inserire e rimuovere

<?php
  $servername = "localhost";
  $username = "xxx";
  $password = "xxx";
  $dbname = "xxx";
  // Create connection
  $conn = new mysqli($servername, $username, $password, $dbname);
  // Check connection
  if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
  }

  $method = $_GET['method'];
  //$user_id = $_GET['user_id'];
  $actor_id = $_GET['actor_id'];

  if ($method == "Like") {
    mysqli_query($conn,"INSERT INTO favs (favoriti_id) VALUES ('$actor_id')");
  }
  else {
    mysqli_query($conn,"DELETE FROM favs WHERE favoriti_id = '$actor_id'");
  }
?>

grazie per la vostra attenzione

[blackskyisback]

aggiornamento, sono riuscito a visualizzare l'icona con qualche modifica al codice php
il problema e' che quando aggiorno la pagina lo stato icona ritorna nullo anche se nel database sono salvati i valori

prima



dopo



dopo refresh pagina

    $actor_id = $_GET["id"];

$favoriti_id = $actor_id;

      function checkFavorite($actor_id, $favoriti_id, $conn) {
        $result = $conn->query("SELECT * FROM favs WHERE actor_id = '". $actor_id."' AND favoriti_id = '". $favoriti_id."'");
        $numrows =  $result->num_rows;
        if ($numrows == 0) {
         echo "<div class = 'button' method = 'Like'  actor_id = ".$actor_id." favoriti_id = ".$favoriti_id."> <img id=".$favoriti_id." src='favoff.png'> </div>";
        }
        else {
          echo  "<div class = 'button' method = 'Unlike'  actor_id = ".$actor_id." favoriti_id = ".$favoriti_id."> <img id=".$favoriti_id." src='favon.jpg'> </div>";
        }
      }
      
      
      
      // Query to get the user_id
      //$result = $conn->query("SELECT * FROM actor WHERE actor_id = $actor_id");
     // $row = $result->fetch_assoc();
     // $actor_id = $row['actor_id'];

      // Query to Get the Director ID
      $result = $conn->query("SELECT * FROM actor WHERE actor_id =" .$actor_id);
      $row = $result->fetch_assoc();
      $actor_id = $row['actor_id'];

     // echo "<p>Actor: ".$row['nome']."</p> ";
      $fav_image = checkFavorite($actor_id, $favoriti_id, $conn);
       echo "Favorite? : ".$fav_image.""; 

[alemoppo]

Non ho controllato bene il codice, ma a occhio mi sembra che la parte interessata sia questa:

$result = $conn->query("SELECT * FROM favs WHERE actor_id = '". $actor_id."' AND favoriti_id = '". $favoriti_id."'");
        $numrows =  $result->num_rows;
        if ($numrows == 0) {
         echo "<div class = 'button' method = 'Like'  actor_id = ".$actor_id." favoriti_id = ".$favoriti_id."> <img id=".$favoriti_id." src='favoff.png'> </div>";
        }
        else {
          echo  "<div class = 'button' method = 'Unlike'  actor_id = ".$actor_id." favoriti_id = ".$favoriti_id."> <img id=".$favoriti_id." src='favon.jpg'> </div>";
        } 

Quindi, sei sicuro che se la query trova 0 risultati, method deve valere 'Like' altrimenti 'Unlike'?

Al caricamento della pagina, appare 'Like'? Ovvero, il codice HTML generato da PHP è corretto? Hai modo di mostrarci una pagina (anche semplificata) che mostra il problema?

Ciao!

[blackskyisback]

Ciao alemoppo

il codice html restituisce questo

	 <!-- FAVORITI TEST INIZIO -->
	 
	 <script src="jquery.min.js"></script>
    <script type="text/javascript">
    jQuery(document).ready(function($){
          $('.button').on('click', function(e){
              e.preventDefault();
              var actor_id = $(this).attr('actor_id'); // Get the parameter user_id from the button
              var favoriti_id = $(this).attr('favoriti_id'); // Get the parameter director_id from the button
              var method = $(this).attr('method');  // Get the parameter method from the button
              if (method == "Like") {
                $(this).attr('method', 'Unlike') // Change the div method attribute to Unlike
                $('#' + favoriti_id).replaceWith('<img class="favicon" id="' + favoriti_id + '" src="favon.jpg">') // Replace the image with the liked button
              } else {
               $(this).attr('method', 'Like')
               $('#' + favoriti_id).replaceWith('<img class="favicon" id="' + favoriti_id + '" src="favoff.png">')
              }
              $.ajax({
                  url: 'favs.php', // Call favs.php to update the database
                  type: 'GET',
                  data: {actor_id: actor_id, favoriti_id: favoriti_id, method: method},
                  cache: false,
                  success: function(data){
                  }
              });
          });
      });
    </script>
	
<div class = 'button' method = 'Like'  actor_id = 40 favoriti_id = 40> <img id=40 src='favoff.png'> </div>Favorite? : 	 
	 
	 
	 
	 
	 <!-- FAVORITI TEST FINE -->

nel database il dato e'ancora presente

[blackskyisback]

risolto

[blackskyisback]

il codice si trova qui, poi ognuno si applica le modifiche in base al suo script e al database

https://github.com/FakeCast/favorite_button