Re[2]: [PHP] imagecolortransparent()

Hi, Tuesday, July 23, 2002, 5:16:43 AM, you wrote: N> "Tom Rogers" <trogers@kwikin.com> wrote in message N> news:16878540805.20020722210409@kwikin.com... >> Hello Nick, >> >> Monday, July 22, 2002, 8:47:39 PM, you wrote: >> >> N> i get php to allocate a colour then make it transparent using >> N> imagecolortransparent() >> >> N> instead of turning it transparent it becomes black. >> >> N> Is it something i'm doing wrong? can anyone help me? >> >> N> thanx in advance >> >> Post a bit of the code so we can see what you are doing >> >> -- >> Tom >> N> $im = imagecreate($size, 21); N> $bgcolour = N> imagecolorallocate($im,hexdec(substr($table[bg],0,2)),hexdec(substr($table[b N> g],2,2)),hexdec(substr($table[bg],4,2))); N> $fontcolour = N> imagecolorallocate($im,hexdec(substr($table[fg],0,2)),hexdec(substr($table[f N> g],2,2)),hexdec(substr($table[fg],4,2))); N> if ($table[bg] == "no") { N> $trans = imagecolortransparent ($im,$bgcolour); N> } probably what you need is: $im = imagecreate($size, 21); if ($table[bg] == "no") { $bgcolour = imagecolorallocate($im,1,1,1); $trans = imagecolortransparent ($im,$bgcolour); } else{ $bgcolour = imagecolorallocate($im,hexdec(substr($table[bg],0,2)),hexdec(substr($table[bg],2,2)),hexdec(substr($table[bg],4,2))); } $fontcolour = imagecolorallocate($im,hexdec(substr($table[fg],0,2)),hexdec(substr($table[fg],2,2)),hexdec(substr($table[fg],4,2))); -- Best regards, Tom