uDesign, # uDesign (28.06.2018 / 17:14)
dayko, Sizga savol
$x = 0;
while($x <= 7) {
$x++;
}
echo $x;
SHu ko`dizda natijasi nechi bo`ladi?8
<?
$c = mysql_query("select * from `tarx` where `user_id` = '".$user_id."' ordery by `id` LIMIT 100");
$i = 0;
while ($d = mysql_fetch_assoc($c)) {
$a = ' != '.$d['savol'].' and ';
$b = mysql_query("select * from `ustun` where $a `mod` = '1'");
echo $b['birbalo'];
}
$i++;
?>
namuna skrenda
skrenda bu köd yöq sababi undan foyqo'shtirnoq "" ichida o'zgaruvchini .. nuqtalar orasiga olishingiz shart emas
namuna skrenda
skrenda bu köd yöq sababi undan foykeyin $a o'zgaruvchida bir mecha marta ' bu belgini yozib kodni chalkashtirib tashagansiz bu hatolik keltirib chiqaradi rasmdagi hatoyingiz shuni ko'rsatayabdi
namuna skrenda
skrenda bu köd yöq sababi undan foybeginner bo'lsangiz aybi yo'q, lekin sql kodlarda xato menimcha, " (ikkita qo'shtirnoq) ishlatsangiz uni ichida faqat ' (bitta qo'shtirnoq) ishlating, sql kod '" . $binnasanom . "' shu yerda yopilib qayta ochilyabi, umuman peremenniylarni sql kod orasida shundoq yozavering ishlayveradi..gif)
.gif)
function newscom($var)
{
return mysql_result(mysql_query("SELECT COUNT(id) FROM news_comm WHERE user = '$var'"), 0);
}
function status($var)
{
$ChEcKeD = newscom($var);
if ($ChEcKeD == 0 || $ChEcKeD <= 15)
{
$status = '<span style="color:#0000FF;">Yangilardan</span>';
$stars = $ChEcKeD;
}
if ($ChEcKeD >= 16 || $ChEcKeD <= 30)
{
$status = '<span style="color:#0000FF;">aktiv</span>';
$stars = $ChEcKeD;
}
if ($ChEcKeD >= 31 || $ChEcKeD <= 50)
{
$status = '<span style="color:#0000FF;">faol</span>';
$stars = $ChEcKeD;
}
elseif ($ChEcKeD >= 51)
{
$status = '<span style="color:#0000FF;">Sayt faxri</span>';
$stars = $ChEcKeD;
}
else
{
$status = NULL;
$stars = NULL;
}
return $status . (isset($stars) ? '<br/>' : '') . $stars . '<br/>';
}