传递的变量在目标页面上的url中返回null

Question

I am going out of my mind as it is. I am trying to pass a variable from a page that shows all albums thumbnail image and name to a page that will display all the pictures in that gallery using that passed variable, but the variable is empty in the url on the target page. I have seen similar cases on the web and on this site and I've applied the suggestions but it's still the same. Here is the code that lists the thumbnail and passes the variable(id).

    <?php
            include ("config.php");
            $conn = mysqli_connect(DB_DSN,DB_USERNAME,DB_PASSWORD,dbname);
            $albums = mysqli_query($conn,"SELECT * FROM albums");
            if (mysqli_num_rows($albums) == 0) {
                echo "You have no album to display. Please upload an album using the form above to get started. ";
            }
            else{
                echo "Albums created so far:<br><br>";
                echo "<table rows = '4'><tr>";
                while ($thumb = mysqli_fetch_array($albums)) {
                    echo '<td><a href ="view.php?id="'.$thumb['id'].'"/><img src = "'.$thumb['thumbnail'].'"/><br>'.$thumb['album_name'].'<br>'.$thumb['id'].'</a></td>';
                }
                echo "</tr></table>";
            }
     ?>

The code for getting the passed variable is as follows:

<?php
include("config.php");
$conn = mysqli_connect(DB_DSN,DB_USERNAME,DB_PASSWORD);
$db = mysqli_select_db($conn,dbname);
if (isset($_GET['id'])) {
    $album_id = $_GET['id'];
    $pic = "SELECT * FROM photos WHERE album_id ='$album_id'";
    $picQuery = mysqli_query($conn,$pic);
    if (!$picQuery) {
        exit();
    }
    if (mysqli_num_rows($picQuery) == 0) {
        echo "Sorry, no Pictures to display for this album";
}
else{
    echo "Pictures in the gallery:<br><br>";
    while ($result = mysqli_fetch_assoc($picQuery)) {
            echo "<img src='".$result['photo_path']."'/>";
        }
    }
}
?>

Please help as i have spent the last two days trying to get it right.

Answer 1

First, your's code is weak against sql injections:

$album_id = $_GET['id']; // here
$pic = "SELECT * FROM photos WHERE album_id ='$album_id'";

Use either $album_id = intval($_GET['id']) or prepared statements functionality.

Second, add debug lines to your code, like:

<?php
include("config.php");
if (isset($_GET['id'])) {
    $album_id = intval($_GET['id']);
    var_dump($album_id); // should print actual passed id

    $conn = mysqli_connect(DB_DSN,DB_USERNAME,DB_PASSWORD);
    var_dump($conn _id); // should print conn resource value

    $db = mysqli_select_db($conn, dbname);
    var_dump($db); // should print 'true' if db select is ok

    $pic = "SELECT * FROM photos WHERE album_id ='$album_id'";
    $picQuery = mysqli_query($conn, $pic);
    var_dump($picQuery); // should print query resource value

    if (!$picQuery) {
        exit();
    }

    if (mysqli_num_rows($picQuery) == 0) {
        echo "Sorry, no Pictures to display for this album";
    } else {
        echo "Pictures in the gallery:<br><br>";
        while (($result = mysqli_fetch_assoc($picQuery)) !== false) {
            var_dump($result ); // should print fetched assoc array
            echo "<img src='".$result['photo_path']."'/>";
        }
    }
}

Notice $album_id = intval($_GET['id']) and while (($result = mysqli_fetch_assoc($picQuery)) !== false) parts

Then follow link view.php?id=<existing-album-id> and observe debug result. On which step debug output differs from expected - there problem is.