xerts
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Posts
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Posts posted by xerts
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nez galigi stupcela
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kur konkreti ?
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Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /www/www.esport.lv/profile/edit/index.php on line 66
66: "SELECT u.user_id,u, user_name, u.user_country.c.country_id,
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es zinu bet kad es izdaru shita
$result = dbquery(
"SELECT u.user_id,u, user_name, u.user_country.c.country_id,
c.country_name, c.country_image
FROM ".PREFIX."users AS u
LEFT JOIN ".PREFIX."country AS c ON (u.user_country=c.country_name)
where u.user_id='$userdata[user_id]'"
);
while ($data = dbarray($result))
{
<img src='/img/flags/".$data['country_image']."'> '".$userdata['user_country']."'</td>
izmet kljudu
}
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$check_url = (isset($_SERVER['REQUEST_URI']) ? $_SERVER['REQUEST_URI'] : $_SERVER['SCRIPT_NAME']);
foreach ($_GET as $check_url) {
if ((eregi("<[^>]*script[^>]*>", $check_url)) || (eregi("<[^>]*object[^>]*>", $check_url)) ||
(eregi("<[^>]*iframe[^>]*>", $check_url)) || (eregi("<[^>]*applet[^>]*>", $check_url)) ||
(eregi("<[^>]*meta[^>]*>", $check_url)) || (eregi("<[^>]*style[^>]*>", $check_url)) ||
(eregi("<[^>]*form[^>]*>", $check_url)) || (eregi("\([^>][^)]*\)", $check_url)) ||
(eregi("<[^>]*frameset[^>]*>", $check_url)) || (eregi("<[^>]*onmouseover[^>]*>", $check_url)) ||
(eregi("<[^>]*img[^>]*>", $check_url)) || (eregi("\"", $check_url)) || (eregi("'", $check_url)))
{
die ();
}
}
un ka jau 4e4en teica htmlspecialchar izmanto
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man vajag ja es ieeju kadam profila .... tad lai liek vinja id un vinja izveleto valsti raksta un parada karodzinu
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izmanto dynamic title
if(@$dynamic_title=="")
{
print $set['sitename'];
}
elseif(@$dynamic_title!="")
{
print @$dynamic_title;
}
tad katra lapa
$dynamic_title = '*****';
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Sveiks
man ir jautajums ka var iznemt datus no vienas db un no otras
piemeram
DB_country
country_id 1
country_name Latvia
country_img lv.gif
DB_users
user_id 1
user_name xerts
user_country 1
Ka man izdarit lai viss sanaktu pareizi un beigu beigas man sanaktu lv karodzinsh un blakus uzrakstita valsts ....
$result = dbquery("SELECT u.user_id, u.user_country, u.user_name,
c.country_id, c.country_name, c.country_img
FROM DB_users AS u
LEFT JOIN DB_country AS c ON (c.country_id=u.user_country)
WHERE .............................. "
while ($data = dbarray($result))
);
palidziet ludzu
mysql
in PHP un PHP :)
Posted
mixis es neesmu jauninajs ... vnk uz visiem veidiem ... man parada syntax error ',' or ';' ......