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Home/ Questions/Q 8101613
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Editorial Team
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Editorial Team
Asked: June 5, 20262026-06-05T23:01:08+00:00 2026-06-05T23:01:08+00:00

Okay so i have this code: if (isset($_GET[‘book’])) { $query= SELECT book_id, title, authors.`author`

  • 0

Okay so i have this code:

if (isset($_GET['book'])) {

$query= "SELECT  book_id, title, authors.`author` , publishers.`publisher` , formats.`format` , genres.`genre` , usersdb.users.username, pages, isbn, description, DATE_FORMAT( books.`release_date` ,  '%M, %d, %Y' ) AS  `release_date` 
    FROM books
    INNER JOIN  `authors` ON (  `authors`.`author_id` =  `books`.`author_id` ) 
    INNER JOIN  `publishers` ON (  `publishers`.`publisher_id` =  `books`.`publisher_id` ) 
    INNER JOIN  `formats` ON (  `formats`.`format_id` =  `books`.`format_id` ) 
    INNER JOIN  `genres` ON (  `genres`.`genre_id` =  `books`.`genre_id` ) 
    INNER JOIN usersdb.users ON ( usersdb.users.user_id = bookdb.books.user_id ) 
    WHERE books.book_id= ".$_GET['book']."";
    $query=@mysql_query ($query, $dbc1);

if ($query) {
var_dump ($query);
while($row=mysql_fetch_assoc($query)){

echo '<h1>'.$row['title'].'</h1>
  <p>'.$row['author'].'</p>
  <p>'.$row['publisher'].'</p>
  <p>'.$row['format'].'</p>
  <p>'.$row['genre'].'</p>
  <p>'.$row['pages'].'</p>
  <p>'.$row['isbn'].'</p>
  <p>'.$row['release_date'].'</p>
  <p>'.$row['description'].'</p>
  <p>'.$row['username'].'</p>';
}
}
}

It works absolutely fine for some of the entries and all the information is displayed correctly and is fine, but for new entries the information doesn’t display at all for some reason.

The mysql query runs correctly and as you can see the I have done a var_dump() and that returns resource(11, mysql result) so the resource is definitely there it just doesn’t seem to get run through the while statement.

If you can see why this is occurring that would be much appreciated.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-05T23:01:09+00:00Added an answer on June 5, 2026 at 11:01 pm

    If you just remove the “@”, you will see that there’s maybe an error. Globally in PHP, the use of “@” is a mistake. You may never use it.

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