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JS iekš PHP.


thesnarkie

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 while($myselect = mysql_fetch_assoc($show_all)) {
				 echo "<td class='titlegreen' align='center'>
					  <table border='0' cellspacing='1' cellpadding='1'>
					  <tr><td style='background: url(http://i.meteorgames.com/ic/bg/000000_both_b48.png) no-repeat;'>

					  <table border='0' cellspacing='0' cellpadding='0'>
					  <tr><td style='background: url(" . $myselect['image'] . ") no-repeat;'>

					  <table border='0' cellspacing='0' cellpadding='0'>  
					  <tr><td><a href='java script: window.open('test.php', 'window_name', 'width = 50, height = 50');'><img src='http://mgimg.net/ic/b/e5e5e5_b48.png' alt='" . $myselect['name'] . "' longdesc='" . $myselect['name'] . "' /></a></td></tr>
					  </table>

					  </td></tr>
					  </table>

					  </td></tr>
					  </table>
					  </td>"; 
					  if(++$i%7==0){
					  echo "</tr><tr>";
						 }
			 }

 

Tur kur tā līnija par pop-up lodziņu - zinu, ka šādi tas nedarbojas, bet kā varētu uztaisīt to, lai darbotos, ja man visa lapa ir.php? Nu es zinu, ka PHP ir servera puses programmēšana un ar to nemaz nevar tādus pop-up logus uztaisīt, bet kā lai pārtrauc lapu pareizi un kas un kā. Vārdu sakot, pasakiet kā es rakstu php skriptu: te. Tas te ir visa lapa un <html><body<head> tagi ir iekš variabliem.

Kā būtu pareizi rakstīt php lapu, lai tajā var izmantot arī JS?

Edited by thesnarkie
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