没有从数据库中选择正确的ID

Question

I am trying to build a CMS system where an admin inserts data in a template. There can be as many pages created with that template as the admin wants. The pages should then be accessed in a menu with links and be displayed with the respective data according the link you just clicked.

Everything is working as expected except the system is not picking up the data from the id that the link had. Instead the system is picking up the data from the last id on the database.

Example: If I click on the link Page#2, I will have the URL http://www.website/page.php?pid=2 but the content that is loaded is the content from the last id on the database and not the 2.

Code to make and display the links:

if (!$_GET['pid']) {
    $pageid = '1';
    } else {
        $pageid = preg_replace('/[^0-9]/', "", $_GET['pid']); // filter everything but numbers for security
    }

    $sqlCommand = "SELECT id, linklabel FROM pages WHERE showing='1' ORDER BY id ASC"; 
    $query = mysqli_query($myConnection, $sqlCommand) or die('Error: ' . mysqli_error($myConnection)); 

    $listadeavatars = '';
    while ($row = mysqli_fetch_array($query)) { 
        $pid = $row["id"];
        $avatar = html_entity_decode($row["linklabel"]);
    } 

    $sqlCommand = "SELECT id, linklabel FROM pages WHERE showing='1' ORDER BY id ASC"; 
    $query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error()); 

    $menuDisplay = '';
    while ($row = mysqli_fetch_array($query)) { 
        $pid = $row["id"];
        $linklabel = $row["linklabel"];
        $avatar = html_entity_decode($row["linklabel"]);

        $menuDisplay .= '<li><a href="page.php?pid=' . $pid . '">' . $avatar . '</a></li>';

    } 
    mysqli_free_result($query);

Code to display the data on page.php according to the ID

if (!$_GET['pid']) {
    $pageid = '1';
} else {
    $pageid = ereg_replace("[^0-9]", "", $_GET['pid']); // filter everything but numbers for security
}

$sqlCommand = "SELECT pagetitle, linklabel, evenemang, presentation, producent, pagebody, mapa FROM pages WHERE id='$pid' LIMIT 1";  
    $query = mysqli_query($myConnection, $sqlCommand) or die (mysqli_error()); 
    while ($row = mysqli_fetch_array($query)) { 
        $pagetitle = $row["pagetitle"];
        $linklabel = $row["linklabel"];
        $evenemang = $row["evenemang"];
        $presentation = $row["presentation"];
        $producent = $row["producent"];
        $pagebody = $row["pagebody"];
        $mapa = $row["mapa"];
        $avatar = html_entity_decode($row["linklabel"]);
    } 
    mysqli_free_result($query);

Please do let me know if I need to explain better! Any suggestions to solve this issue? Cheers!

Answer 1

On page.php, change WHERE id='$pid'

$sqlCommand = "SELECT pagetitle, linklabel, evenemang, presentation, producent, pagebody, mapa FROM pages WHERE id='$pid' LIMIT 1"; 

to WHERE id='$pageid'

$sqlCommand = "SELECT pagetitle, linklabel, evenemang, presentation, producent, pagebody, mapa FROM pages WHERE id='$pageid' LIMIT 1";