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(Solved) Ways to run the join command on memory/variable contents instead of files? (programming.dev)

submitted 17 hours ago* (last edited 16 hours ago) by loveknight@programming.dev to c/shell@programming.dev

3 comments fedilink hide all child comments

Edit: I figured it out: The solution is over a process substitution operator:

join <(echo "$var1") <(echo "$var2")

See also the comment by @notabot@piefed.social below, this StackOverflow comment, and the GNU documentation..

It's comparatively straightforward to use the content of one variable in join by using - to tell it to use standard input for that file:

echo $variable | join - anotherfile

However, is there a way to serve both input 'files' from variables, avoiding temporary files on the disk?

It seems like the easiest way would be via creating and mounting a temporary partition via tmpfs,

mount -t tmpfs -o size=50m tmpfs /mountpoint,

and just create temporary files in there. And afterwards clean things up.

So far I've also attempted here-documents, but apparently this too can only provide standard input, so that the other input still has to be served from a file.

Maybe one can also try doing it via named pipes (mkfifo), but I fear this could introduce lots of potentialities for errors.

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[–] notabot@piefed.social 2 points 17 hours ago (2 children)

You can use the <( ) replacement for this as it runs a command in a subshell and replaces it with the filename of a temporary fifo. So something like:

join <( echo "${var1}" ) <( echo "${var2}" )

[–] loveknight@programming.dev 1 points 16 hours ago* (last edited 16 hours ago) (1 children)

Perfect, thanks for the explanation. Indeed, I found the same solution via StackOverflow about simultaneously.

[–] notabot@piefed.social 1 points 16 hours ago* (last edited 16 hours ago)

The bash man page is full of useful bits like this, but is is densely packed. It's definitely worth reading though.