'How to get script directory in POSIX sh?

I have the following code in my bash script. Now I wanna use it in POSIX sh. How can I convert it?

DIR="$( cd "$( dirname "${BASH_SOURCE[0]}" )" > /dev/null && pwd )"
bashposixsh


Solution 1:[1]

@City responded that 

DIR=$( cd -P -- "$(dirname -- "$(command -v -- "$0")")" && pwd -P )

works. I used it too.
I found the command at https://stackoverflow.com/questions/760110/can-i-get-the-absolute-path-to-the-cu??rrent-script-in-kornshell.

Solution 2:[2]

if      OLDPWD=/dev/fd/0 \
        cd - && ls -lLidFH ?
then    cd . <8
fi      </proc/self/fd 8<. 9<$0

there. that should enable you to change directpry through some magic links as file descriptors.

setting $OLDPWD pre-cd exports the value for the duration of the one change directory (note: cd can have residual effects on hash tables, but the only sh about which i am aware that actually males any good use of these is kevin ahlmquists - and since herbert xu - dash, and maybe some bsd stuff, but what do i know?) but does not carry over any cd exports as a result of the change.

thus, $OLDPWD does not change, actually, and if it had any value at all this remains as was. $PWD is changed as a result of the first cd, and the value becomes /dev/fd/0 which points to /proc/self/fd, where a list of file descriptors for our process should be in ., to include whatever $0 is on ./2.

so we do an ls ... ? and have a look at all of the wonderful info we can get, and we go whence we came.

yay!

Sources

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Source: Stack Overflow

Solution Source
Solution 1 Community
Solution 2 mikeserv

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