將php值解析為已點燃的數據表

Question

希望還有其他人正在使用經過交叉的點火數據表...我已經嘗試了一段時間了，為什么它根本不讓我使用PHP值沒有意義->

include $DOCUMENT_ROOT.'/include/config.inc';
include $DOCUMENT_ROOT.'/include/session_check.inc';

require_once($DOCUMENT_ROOT.'/include/Datatables.php');
$datatables = new Datatables();

// MYSQL configuration
$config = array(
'username' => $dbuser,
'password' => $dbpass,
'database' => $dbname,
'hostname' => $dbhost);

$datatables->connect($config);

$datatables
->select('users.id as user_id, users.name as realname, users.username as username, users.email as email, users.phone as phone, domain.name as domain_name')
->from('users')
->join('domain', 'domain.id = users.domain_id', 'left')
->where('users.domain_id = "$domain_id"')
->add_column('available', '$1', 'available(available)')
->add_column('edit', '<a href="/wh.php?edit=1&id=$1" title="Purge the info for this Stock"><img src="/images/icons/dark/pencil.png" border="0"></a>', 'stock_id')
->unset_column('user_id');

echo $datatables->generate();

現在，值$ domain_id完全不可用。 我想知道我是否錯誤地輸入了它，或者在這樣的字符串組成中可以接受？

Answer 1

好吧..我的錯誤..而不是線

->where('users.domain_id = "$domain_id"')

我應該用這個

->where('users.domain_id', $domain_id)

它現在一切都很好.. thanx :-)