Gunchaz Posted January 19, 2005 Report Posted January 19, 2005 <? $filename = 'poga.gif'; $degrees = 18; header('Content-type: image/jpeg'); $source = imagecreatefromjpeg($filename); $rotate = imagerotate($source, $degrees, 0); imagejpeg($rotate); ?> izvadaas bildiite skiibi !!! Zuper duper. Bet vajag taa lai taa buutu podzinja, un vajag lai nospiezot lido uz kaukaadu linku. Jautaajums taads kaa pieskirt linku bildiitei tipa mainiigajam $rotate ?!
рпр Posted January 19, 2005 Report Posted January 19, 2005 tā nevar. to var ar htmlu izdariit, iekhs <a href="links.html"><img src="fails.php"/></a> jānorāda php skripts (fails.php), kas to bildi parādīs. ja vajag podzinja, tad <input type="image" src="fails.php"/>
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