用regex -php操纵img src

Question

我该如何替换它：

http://example.com/myfolder/files/year/month/imagename-widthxheight.imageextension

对此：

http://example.com/myfolder/files/year/month/imagename-300x300.imageextension

任何帮助？

Answer 1

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");

实际值-300x300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。

Answer 2

怎么样：

$new_img = preg_replace("~([^/]+)-widthxheight(\.[^.]+)$~i","$1-300x300$2", 
"http://www.mysite.com/myfolder/files/year/month/imagename-widthxheight.imageextension");