用于在PHP中过滤文件名的正则表达式

Question

I have strings that represent file paths. Something like this.

folder1/folderA/file1.flv  
folder2/folderB/file1.mp4  
folder3/folderC/file1.jpg

I am trying to write an expression that basically matches against a list of valid extensions such as .flv, .mp4, .flv

I am sure I am way off here but here's what I have .[(\.flv/)|(\.wmv/)|(\.mp4)] comical, I'm sure... but hopefully shows the gist.

Answer 1

If I interpret it correctly, you probably just want something like:

/[.](flv|mp4|wmv)$/

The $ ensures that it matches at the end of the string/filename. The alternatives are listed in the parens, and the [.] is just a nicer notation for \.