Extract filename and path from URL in bash script

Question

In my bash script I need to extract just the path from the given URL. For example, from the variable containing string:

http://login:[e-mail]/one/more/dir/file.exe?a=sth&b=sth

I want to extract to some other variable only the:

/one/more/dir/file.exe

part. Of course login, password, filename and parameters are optional.

Since I am new to sed and awk I ask you for help. Please, advice me how to do it. Thank you!

Answer 1

Using only bash builtins:

path="/${url#*://*/}" && [[ "/${url}" == "${path}" ]] && path="/"

What this does is:

1. remove the prefix *://*/ (so this would be your protocol and hostname+port)
2. check if we actually succeeded in removing anything - if not, then this implies there was no third slash (assuming this is a well-formed URL)
3. if there was no third slash, then the path is just /

note: the quotation marks aren't actually needed here, but I find it easier to read with them in