Parašė hexon· 2011 Rugp. 23 17:08:11
#4
<?php
$xxx = $xx - 2;
$yyy = $yy + 2;
$x2 = $xx + 2;
$y2 = $yy - 2;
$kor = mysql_query("SELECT * FROM koordinates WHERE (x>='$xxx' and y>='$yyy') and (x<='$x2' and y<='$y2') LIMIT 0, 25");
$m = 0;
while($kord = mysql_fetch_array($kor))
{
$m = $m + 1;
if($kord[kieno] != "none"){$gof = "images/map/Gyvenviete.gif";}
if($kord[kieno] == "none"){$gof = "images/map/laukas.gif";}
?>
<table width="0" height="0" cellspacing="0" cellpadding="0" border="0">
<tr>
<a class='urlz' href="map.php?id=aboutvillage&kieno=$kor[uid]"><img class="ico" src="<?echo"$gof";?>" alt="MAP" title=""/></a>
</tr>
</table>
<?
}
?>
Dar viena problema :DDD Gaunu vertikalia eile kurioje visi 25 paveiksleliai. Kaip man padaryti 5X5??? ?|
Redagavo hexon· 2011 Rugp. 23 18:08:23