Ok so I have one page, called albums.php, with the following code:
<?php
require_once('./config.php');
$mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
$album = $mysqli->query("SELECT * FROM Albums");
print('<div id="grid">');
print('<ul>');
while ($row = $album->fetch_assoc()) {
$cover = $row['Album_Cover'];
$name = $row['Album_Name'];
$id = $row['Album_ID'];
print ('<li>');
print('<form method="POST" action="">');
print("<input type='image' src=$cover name='image' id='image' class=$id>");
print("</form>");
print('<br/>');
print ("$name");
print ('</li>');
}
print('</ul>');
print('</div>');
print('<br/>');
print('<br/>');
print('<br/>');
print('<br/>');
$mysqli->close();
?>
DB_HOST, DB_USER, DB_PASSWORD, and DB_NAME come from config.php. albums.php (again, the code above) is linked to the script button.js, the code for which is below:
$("#image").click(function(e) {
e.preventDefault();
var id = $(this).attr('class');
var dataString = 'id='+id;
$.ajax({
type: "POST",
data: dataString,
url: "./pictures.php",
success: function(data) {
alert(data);
}
});
});
My goal is to pass the id of the clicked image to pictures.php using ajax. My code for pictures.php is as follows:
<?php
require_once('./config.php');
require_once('./albums.php');
$id = $_POST['id'];
$mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
$pictures = $mysqli->query("SELECT * FROM Pictures WHERE Album_ID = $id");
print('<div id="grid">');
print('<ul>');
while ($row = $pictures->fetch_assoc()) {
$picture = $row['Picture'];
print("<li><img src=$picture class='thumbnail'></li>");
}
print('</ul>');
print('</div>');
$mysqli->close();
?>
Do I also have to link the button.js script in pictures.php? Other than that, I can't think of a possible issue with this code. By the way, all three of these files are stored in the same folder on my server, so I believe I am accessing them correctly. Any help would be much appreciated!
