Should PATH contain directories or full paths to binaries?

I am trying to set up a correct PATH, but I'm wondering what it should contain. If I have

/usr/bin/ls
/usr/local/bin/ls

and I want to prefer the one in /usr/local/bin, which of the following should I use?

PATH=/usr/local/bin/ls:/usr/bin/ls
PATH=/usr/local/bin:/usr/bin

or something else entirely?

This is not per se a suitable question for Stack Overflow. I expect this to be closed as General Computing or Too Broad; but the answer is frequently needed by beginners, so I hope this won't be deleted.

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 …

1. … look for an executable with the path /usr/local/bin/ls/ls (note the double ls at the end).

2. 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.

3. 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.

Indeed, as you can easily find out through experimentation, the variable PATH should already contain a list of directories which are consulted in that order. In fact, you should already find that you have /usr/local/bin and /usr/bin in the default PATH, usually in this order (though perhaps with other directories between them, and certainly with more directories around them).

To inspect your current PATH value, try

echo "$PATH"

or for a slightly more human-readable rendering

echo "${PATH//:/$'\n'}"     # bash only

or

echo "$PATH" | tr ':' '\012'  # any POSIX system

If you managed to set your PATH to an invalid value (which would cause simple commands like ls and cat to no longer be found, and produce command not found errors) you can try

PATH=/usr/local/bin:/usr/bin:/usr

to hopefully at least restore the essential functionality so that you can use cp or a simple system editor to get back to the original, safe, system default PATH.