Искарване на имена от датабаза

[lqlqlq]

<?php
$connection = mysql_connect($hostname, $user, $pass);        
if(!$connection) { 
die(mysql_error());      
} 
mysql_select_db("$database", $connection) or die(mysql_error()); 
$query = mysql_query("SELECT guid, name, race, class, gender, level, zone FROM characters WHERE online = '1' AND (NOT 'extra_flags' & 1 AND NOT 'extra_flags' & 16) ORDER BY `name`", $connection); 
$total = mysql_num_rows($query); 
echo $total;  ?>

Това е скрипта които извлича уж имената, но ги искарва като цифри, аз искам да ги искарва като имена, колоната на имената е "name"
Как трябва да стане?

 

[oveRLuckEd]

<?php
$connection = mysql_connect($hostname, $user, $pass);
if(!$connection) {
die(mysql_error());
}
mysql_select_db("$database", $connection) or die(mysql_error());
$query = mysql_query("SELECT guid, name, race, class, gender, level, zone FROM characters WHERE online = '1' AND (NOT 'extra_flags' & 1 AND NOT 'extra_flags' & 16) ORDER BY `name`", $connection);
$total = mysql_fetch_assoc($query);
echo $total; ?>

 

[lqlqlq]

нестава..пише само 1 "Array"

 

[Karcho]

<?php
$connection = mysql_connect($hostname, $user, $pass);
if(!$connection) {
die(mysql_error());
}
mysql_select_db("$database", $connection) or die(mysql_error());
$query = mysql_query("SELECT guid, name, race, class, gender, level, zone FROM characters WHERE online = '1' AND (NOT 'extra_flags' & 1 AND NOT 'extra_flags' & 16) ORDER BY `name`", $connection);
$total = mysql_fetch_assoc($query);
echo $total['name']; ?>

 

[lqlqlq]

<?php
$connection = mysql_connect($hostname, $user, $pass);
if(!$connection) {
die(mysql_error());
}
mysql_select_db("$database", $connection) or die(mysql_error());
$query = mysql_query("SELECT guid, name, race, class, gender, level, zone FROM characters WHERE online = '1' AND (NOT 'extra_flags' & 1 AND NOT 'extra_flags' & 16) ORDER BY `name`", $connection);
$total = mysql_fetch_assoc($query);
echo $total['name']; ?>

Да, така става, само ми е чудно защо искарва само 1, като са на линия 4-ма ?

 

[sharky_design]

online = '1'

Заради това...

 

[lqlqlq]

online = '1'

Заради това...

Не мисля, точно това е колоната в която се отразява дали е онлайн дадения герой, ако е 0 тогава ще стане много лошо при 900 акаунта..

ПП: при 0 пак показва само 1 име :lol:

 

[Karcho]

<?php
$connection = mysql_connect($hostname, $user, $pass);
if(!$connection) {
die(mysql_error());
}
mysql_select_db("$database", $connection) or die(mysql_error());
$query = mysql_query("SELECT guid, name, race, class, gender, level, zone FROM characters WHERE online = '1' AND (NOT 'extra_flags' & 1 AND NOT 'extra_flags' & 16) ORDER BY `name`", $connection);
while($total = mysql_fetch_assoc($query)) {
echo $total['name'];
}
?>

така и си готов

//shark, как ще е от това

 

[lqlqlq]

<?php
$connection = mysql_connect($hostname, $user, $pass);
if(!$connection) {
die(mysql_error());
}
mysql_select_db("$database", $connection) or die(mysql_error());
$query = mysql_query("SELECT guid, name, race, class, gender, level, zone FROM characters WHERE online = '1' AND (NOT 'extra_flags' & 1 AND NOT 'extra_flags' & 16) ORDER BY `name`", $connection);
while($total = mysql_fetch_assoc($query)) {
echo $total['name'];
}
?>

Karcho, страшен си обаче, ще те помоля за една екстра, незнам дали ще можеш да я направиш да има <br /> след всеки ник, незнам как трябва да стане.. с echo ли..

 

[Karcho]

<?php
$connection = mysql_connect($hostname, $user, $pass);
if(!$connection) {
die(mysql_error());
}
mysql_select_db("$database", $connection) or die(mysql_error());
$query = mysql_query("SELECT guid, name, race, class, gender, level, zone FROM characters WHERE online = '1' AND (NOT 'extra_flags' & 1 AND NOT 'extra_flags' & 16) ORDER BY `name`", $connection);
while($total = mysql_fetch_assoc($query)) {
echo $total['name'].'<br/>';
}
?>

 

[lqlqlq]

+1