Lancom Posted July 19, 2008 Report Share Posted July 19, 2008 (edited) Кто узнант как зделать типа http://youweb.com/("?pg=") Edited July 19, 2008 by Lancom Link to comment Share on other sites More sharing options...
Kiwix Posted July 19, 2008 Report Share Posted July 19, 2008 Ko Tu gribi uztaisīt??? Link to comment Share on other sites More sharing options...
Lancom Posted July 19, 2008 Author Report Share Posted July 19, 2008 es gribu lai beigas budut ?pg=download . td Link to comment Share on other sites More sharing options...
*jancis38* Posted July 19, 2008 Report Share Posted July 19, 2008 PHP <?php switch($_GET['pg']) { case 'downloads': include( "downloads.php"); break; default: echo "<a href='?pg=downloads'>downloads</a>"; break; } ?> Link to comment Share on other sites More sharing options...
Lancom Posted July 19, 2008 Author Report Share Posted July 19, 2008 (edited) я просто у себя зделал дизайн а если я хачу чтоб у кождого bila ?pg=forum ?pg=shop <a href="?pg=Shop">link</a><br> Edited July 19, 2008 by Lancom Link to comment Share on other sites More sharing options...
*jancis38* Posted July 19, 2008 Report Share Posted July 19, 2008 (edited) Tebe nado lažit etot kod, gde ti evo ho4e6 videt. !DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> ... </head> <body> header <?php switch($_GET['pg']) { case 'downloads': include("downloads.php"); break; case 'files': include("files.php"); break; case 'news': include("news.php"); break; default: echo "<a href='?pg=downloads'>downloads</a>"; break; } ?> footer </body> </html> Nu i tak laži etih case i vse. EDIT: Nu vot naprimer ti napisal ?pg=files i on tebe vivedit include("files.php"); Edited July 19, 2008 by *jancis38* Link to comment Share on other sites More sharing options...
Lancom Posted July 19, 2008 Author Report Share Posted July 19, 2008 (edited) можеш обеснить здесь и за чего неидет mysql если он включен Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in c:\appserv\www\access.php on line 39 ета у меня на sms acc Жду ответа!! Edited July 19, 2008 by Lancom Link to comment Share on other sites More sharing options...
Lancom Posted July 19, 2008 Author Report Share Posted July 19, 2008 Жду ответа!! Link to comment Share on other sites More sharing options...
*jancis38* Posted July 19, 2008 Report Share Posted July 19, 2008 access.php pokozhi. Link to comment Share on other sites More sharing options...
Lancom Posted July 19, 2008 Author Report Share Posted July 19, 2008 я в приват чтоб нескопировали Link to comment Share on other sites More sharing options...
forcex Posted July 22, 2008 Report Share Posted July 22, 2008 neticu Link to comment Share on other sites More sharing options...
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