защо не ми изкарва заявките?

[stun4o]

.. цялото echo не работи ?!

<?php
include ("config.php");
if(isset($_GET['id'])){
$id = (int)$_GET['id'];
$query1 = "SELECT * FROM `bans` ORDER BY id";
$result1=mysql_query($query1);
while($r = mysql_fetch_array($result1)) {
extract($r);
}
echo "
$ip - $reason <br />
";
}
?>

 

[nogravity]

<?php
include ("config.php");
if(isset($_GET['id'])){
$id = (int)$_GET['id'];
$query1 = "SELECT * FROM `bans` ORDER BY id";
$result1=mysql_query($query1);
while($r = mysql_fetch_array($result1)) {
extract($r);
}
echo $ip - $reason "<br />";
}
?>

Така?

 

[alex95_bg_2]

 <?php
include ("config.php");
if(isset($_GET['id'])){
$id = (int)$_GET['id'];
$query1 = "SELECT * FROM `bans` ORDER BY id";
$result1=mysql_query($query1) or die(mysql_error());
while($r = mysql_fetch_array($result1)) {
extract($r);
}
echo "
$ip - $reason <br />
";
}
?>

 

[stun4o]

дава ми грешка
Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting ',' or ';' in C:\xampp\htdocs\bans\admin.php on line 10

,а ако ограда това в "" няма да ми ги извади :/

 

[stun4o]

alex_95... не става

 

[Anonymous]

<?php
include ("config.php");
if(isset($_GET['id'])){
$id = (int)$_GET['id'];
$query1 = "SELECT * FROM `bans` ORDER BY id";
$result1=mysql_query($query1);
while($r = mysql_fetch_array($result1)) {
echo "
$r[ip] - $r[reason] <br />
";
}
?>

 

[stun4o]

мм No

 

[nogravity]

<?php
include ("config.php");
if(isset($_GET['id'])){
$id = (int)$_GET['id'];
$query1 = "SELECT * FROM `bans` ORDER BY id";
$result1=mysql_query($query1) or die(mysql_error());
while($r = mysql_fetch_array($result1)) {
echo $r['ip'] - $r['reason']. "<br />";
}
?>

 

[stun4o]

пак не става
дава ми $end на 10-ти ред .. слагам }
и нищо. ..

 

[stun4o]

:|

 

[HeLL_OwNeR]

 <?php
include ("config.php");
if(isset($_GET['id'])){
$id = (int)$_GET['id'];
$query1 = "SELECT * FROM `bans` ORDER BY id";
$result1=mysql_query($query1) or die(mysql_error());
while($r = mysql_fetch_array($result1)) {
extract($r);
}
echo "
['.$ip.'] - ['.$reason.'] <br />
";
}
?>

Така става ли ?

 

[stun4o]

не става.. в 5 реда код една грешка не мойм открием

 

[alex95_bg_2]

<?php
include ("config.php");
if(isset($_GET['id'])){
$id = (int)$_GET['id'];
$query1 = "SELECT * FROM `bans` ORDER BY id";
$result1=mysql_query($query1) or die(mysql_error());

while($r = mysql_fetch_array($result1)) {
extract($r);

echo "
$ip - $reason <br />
";

}

?>

както беше ще ти показва последния бан само
сега?

 

[stun4o]

аре този път няма да барам нищо..

Parse error: syntax error, unexpected $end in C:\xampp\htdocs\bans\admin.php on line 16

:?

 

[alex95_bg_2]

 <?php
include ("config.php");

$query1 = "SELECT * FROM `bans` ORDER BY id";
$result1 = mysql_query($query1) or die(mysql_error());

while($r = mysql_fetch_array($result1)) {
extract($r);

echo "
$ip - $reason <br />
";

}

?>

 

[stun4o]

бял екран.. (а и аз така пробвах..)

 

[alex95_bg_2]

  <?php
include ("config.php");

$query1 = "SELECT * FROM `bans` ORDER BY id";
$result1 = mysql_query($query1) or die(mysql_error());

echo "<pre>";

while($r = mysql_fetch_assoc($result1)) {

print_r($r);

echo "\n\n\n";

}

echo "</pre>";


?>

запази си стария код, но копирай каквото изкарва този

а и едитнах си предното мнение, пробвай и с другия код :

 <?php
include ("config.php");

$query1 = "SELECT * FROM `bans` ORDER BY id";
$result1 = mysql_query($query1) or die(mysql_error());

while($r = mysql_fetch_array($result1)) {
extract($r);

echo "
$ip - $reason <br />
";

}

?>

 

[stun4o]

Array
(
[id] => 14
[ip] => 91.201.175.3
[reason] => taka iskam!
)

ми на това как да му сложа "изтрии" "Редактирай"

 

[alex95_bg_2]

ok.. това беше за пробва

а кода отдолу работи ли? втория

 

[stun4o]

ПЕРФ +1