我该如何替换这个:
http://example.com/myfolder/files/year/month/imagename-widthxheight.imageextension
到此:
http://example.com/myfolder/files/year/month/imagename-300x300.imageextension
有什么帮助吗?
你可能想试试这个
// if your src has widthxheight are specified literally like that you may try
echo preg_replace("/'W{0,1}(width).*(height)/i","-300x300","http://www.mysite.com/myfolder/files/year/month/imagename-widthxheight.imageextension");
// if your src has widthxheight are specified in int val you may try
echo preg_replace("/'W{0,1}('d{1,7}).*('d{1,7})/i","-300x300","http://www.mysite.com/myfolder/files/year/month/imagename-123x456.imageextension");
-300x300的实际值将根据您的实际需求而有所不同。所以我认为最好通过变量传递这些值。
怎么样:
$new_img = preg_replace("~([^/]+)-widthxheight('.[^.]+)$~i","$1-300x300$2",
"http://www.mysite.com/myfolder/files/year/month/imagename-widthxheight.imageextension");