PATH
works only with directories, not with single files
From the POSIX standard (emphasis mine)
PATH
This variable shall represent the sequence of path prefixes that certain functions and utilities apply in searching for an executable file known only by a filename. The prefixes shall be separated by a colon ( ':' ). [...] The list shall be searched from beginning to end, applying the filename to each prefix, until an executable file with the specified name and appropriate execution permissions is found.
When you type in ls
into your shell and your PATH
is set to /usr/local/bin/ls:/usr/bin/ls
then your shell will …
… look for an executable with the path /usr/local/bin/ls/ls
(note the double ls
at the end).
As that path does not exist on your system your shell will proceed to look for an executable with the path /usr/bin/ls/ls
(double ls
again). That path also doesn't exist.
The shell couldn't find an executable using all paths in PATH
so your shell will print something like bash: ls: command not found
.
So, what did we learn? Paths listed by PATH
have to be directories. You cannot list single files. Therefore the correct answer in your case is PATH=/usr/local/bin:/usr/bin
.
Where things get interesting
Imagine the following situation. You have two versions of program c1
and two versions of program c2
. The different versions are stored in the directories /a/
and /b/
.
/a/c1
/a/c2
/b/c1
/b/c2
How can we set PATH
to prefer /a/c1
over /b/c1/
but at the same time /b/c2
over /a/c2
?
Sadly, there is no way to achieve this directly as we can only specify directories in PATH
. We have to move/rename some files or create symlinks and use the symlinks inside the paths. One possible solution:
mkdir /c
ln -s /a/c1 /c/c1
ln -s /b/c2 /c/c2
export PATH=/c:/a:/b
The trailing :/a:/b
is not really necessary here. I included them under the assumption that /a
and /b
contain more executables than just c1
and c2
.
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