使用sed和regex捕获url的最后一部分

Question

I'm trying to make sed match the last part of a url and output just that. For example:

echo "http://randomurl/suburl/file.mp3" | sed (expression)

should give the output:

file.mp3

So far I've tried sed 's|\\([^/]+mp3\\)$|\\1|g' but it just outputs the whole url. Maybe there's something I'm not seeing here but anyways, help would be much appreciated!

Answer 1

这工作：

 echo "http://randomurl/suburl/file.mp3" | sed 's#.*/##'

Answer 2

basename is your good friend.

> basename "http://randomurl/suburl/file.mp3"
=> file.mp3

Answer 3

This should do the job:

$ echo "http://randomurl/suburl/file.mp3" | sed -r 's|.*/(.*)$|\1|'
file.mp3

where:

• | has been used instead of / to separate the arguments of the s command.
• Everything is matched and replaced with whatever if found after the last / .

Edit: You could also use bash parameter substitution capabilities:

$ url="http://randomurl/suburl/file.mp3"
$ echo ${url##*/}
file.mp3

Answer 4

echo 'http://randomurl/suburl/file.mp3' | grep -oP '[^/\n]+$'

这是使用grep的另一种解决方案。