servera monitors

[homers]

Tatad gribu izvilkt lietotaja profilu monitorā, jā spēletaja niks ir reģistrets iekš forumā uz viņu rada linku, bet jā nav vienkarši rada spēletaju niku bez linka.

Rekur kaut kads kods man ir, bet viņš neiet vienkarši rada niku bez linka:

$name=SubStr($st,0,StrPos($st,chr(0)));
$a=  mysql_fetch_row(mysql_query("SELECT name FROM ibf_members"));
$name_f = $a[0];
$af=  mysql_fetch_row(mysql_query("SELECT id FROM ibf_members WHERE name=$name_f"));
$name_f_id = $af[0];
if($name == $name_f){
echo "<a href='?q=user&id=".$name_f_id."'>$name</a>";
}
else
{
echo "$name";
}

[EdgarsK]

<?php
$name=substr($st,0,strpos($st,chr(0)));
$check=mysql_query("select id from ibf_members where name='$name'");
if(mysql_num_rows($check)) {
$names=mysql_fetch_array($check);
echo "<a href='?q=user&id=".$names["id"]."'>$name</a>";
}else echo $name;

? shaadi darbotos ?

Edited September 27, 2009 by EdgarsA

[homers]

Nē, tapat rada spēletaja niku bez linka

[marcis]

Meklē tepat forumā, atceros, ka arī šāda tēmas modifikācija ir pārrunāta.

[cunamiX]

Pamēģini šādi

<?php
$name=substr($st,0,strpos($st,chr(0)));
$check=mysql_query("SELECT id FROM ibf_members WHERE members_display_name = '$name'");
$check2=mysql_fetch_assoc($check);
if(isset ($check2['id'])){
$names .="<a href=http://links/?q=user&id={$names['id']}>$name</a>";
}else{
$names .="$name";
}

Edited September 27, 2009 by cunamiX

[homers]

EdgarsA jau palidzeja, paldies visiem