Помощь по скриптам | Не выводится полный список
Тема закрыта by
282326970
Причина: 1
Причина: 1
Почему выводит только первую строку ?
$resultat = mysql_query("SELECT * FROM `Logs` WHERE `id`",$db);
$array = mysql_fetch_array($resultat);
echo "<tr><td>#".$array['id']."</td><td>".$array['admin']."</td><td>".$array['adminip']."</td><td>".$array['player']."</td><td>".$array['action']."</td><td>".$array['amount']."</td><td>".$array['reason']."</td></tr>
</table>";
$resultat = mysql_query("SELECT * FROM `Logs` WHERE `id`",$db);
$array = mysql_fetch_array($resultat);
echo "<tr><td>#".$array['id']."</td><td>".$array['admin']."</td><td>".$array['adminip']."</td><td>".$array['player']."</td><td>".$array['action']."</td><td>".$array['amount']."</td><td>".$array['reason']."</td></tr>
</table>";
DELETED
3 июля 2017, в 16:54
Delete
shadrvlad , с таблицы logs всю информацию
$resultat = mysql_query("SELECT * FROM `Logs` WHERE `id`",$db);
while($result = mysql_fetch_array($resultat)){
echo "<tr><td>#".$result['id']."</td><td>".$result['admin']."</td><td>".$result['adminip']."</td><td>".$result['player']."</td><td>".$result['action']."</td><td>".$result['amount']."</td><td>".$result['reason']."</td></tr>
</table>";
}
//
while($result = mysql_fetch_array($resultat)){
echo "<tr><td>#".$result['id']."</td><td>".$result['admin']."</td><td>".$result['adminip']."</td><td>".$result['player']."</td><td>".$result['action']."</td><td>".$result['amount']."</td><td>".$result['reason']."</td></tr>
</table>";
}
//
DELETED
3 июля 2017, в 16:58
Delete
shadrvlad , 7 пост
282326970 , лол, сек .
282326970 , пробуй так
$res = mysql_query("SELECT * FROM `Logs` ORDER BY `id`"$db);
while($result = mysql_fetch_array($res)) {
echo "<tr><td>#".$result['id']."</td><td>".$result['admin']."</td><td>".$result['adminip']."</td><td>".$result['player']."</td><td>".$result['action']."</td><td>".$result['amount']."</td><td>".$result['reason']."</td></tr>
</table>";
} ///
Или
$result = mysql_fetch_assoc(mysql_query("SELECT * FROM `logs` where `id`"$db));
echo "<tr><td>#".$result['id']."</td><td>".$result['admin']."</td><td>".$result['adminip']."</td><td>".$result['player']."</td><td>".$result['action']."</td><td>".$result['amount']."</td><td>".$result['reason']."</td></tr>
</table>";
?>
$res = mysql_query("SELECT * FROM `Logs` ORDER BY `id`"$db);
while($result = mysql_fetch_array($res)) {
echo "<tr><td>#".$result['id']."</td><td>".$result['admin']."</td><td>".$result['adminip']."</td><td>".$result['player']."</td><td>".$result['action']."</td><td>".$result['amount']."</td><td>".$result['reason']."</td></tr>
</table>";
} ///
Или
$result = mysql_fetch_assoc(mysql_query("SELECT * FROM `logs` where `id`"$db));
echo "<tr><td>#".$result['id']."</td><td>".$result['admin']."</td><td>".$result['adminip']."</td><td>".$result['player']."</td><td>".$result['action']."</td><td>".$result['amount']."</td><td>".$result['reason']."</td></tr>
</table>";
?>
Стр.: 1, 2