Find path in bash on insensitive manner

Question

Suppose a path like

/home/albfan/Projects/InSaNEWEBproJECT

Despite of the fact to not use such that names. Is there a way to check for a path

Answer 1

The simplest solution:

$ find . | grep -qi /path/to/something[^/]*$ 

But if you have some additional conditions that must be checked for matched file, you can run grep inside find:

$ find . -exec sh -c 'echo {} | grep -qi /path/to/something' \; -print

Here you will get all files that are in the directory. If you want to get only the directory's name:

$ find . -exec sh -c 'echo {} | grep -qi /path/to/something[^/]*$' \; -print

Example of usage:

$ mkdir -p Projects/InSaNEWEBproJECT/src/main/resources/
$ find . -exec sh -c 'echo {} | grep -qi /projects/insanewebproject[^/]*$' \; -print
./Projects/InSaNEWEBproJECT

Answer 2

My fault! I guess I tried

find -ipath 'projects/insanewebproject' 

but the trick here is that I must use

find -ipath './projects/insanewebproject'

That ./ does the change. Thanks!.

man says -path is more portable than -wholename

if you expect only one result, you can add | head -n1, cause that way head kill pipe when it fills its buffer, which is only one line length

find -ipath './projects/insanewebproject'| head -n1