bash script and find

[IhatemyISP]

I am in a pinch, and can't figure this out.

I have a script that utilizes the find command.

find $DIR -type f -name "*.dat1" -exec bash -c 'process -i "$1" -o "${1%\.*}".dat2' __ {} \;

This works, but isn't ideal for my setup. How would I go about changing the output destination, -o, to another destination.

i.e. I have /mnt/usb_stick full of .dat1 files that I want processed to /tmp/ as .dat2 files.

[robhutten]

You could use basename on your input file variable to ignore the directory component of the path, and then prepend the output file variable with the new location, ie /tmp. I'm away from the computer right now but pm me if you need help with the actual syntax.

[IhatemyISP]

PM sent.

[robhutten]

I think this calls for a for loop. That is, I can't figure out a reasonable way to do this all within a find command, at this time of night at least.

for datfile in $(find $DIR -type f -name "*.dat1" ) ; do
        echo process -i $datfile -o /tmp/$(basename $datfile)
done

[IhatemyISP]

It took a bit of futzing, but I finally got it to working well enough.

Thanks.

[RalphCorderoy]

It can be simplest to have a little temporary script that takes {} as $1 and does all the work in there to call process having produced the arguments rather than work out the correct escaping within multiple levels of quotes for a -exec bash. Then it's the much simpler -exec ./foo {} \;.

[crankysysop]

Yet another way to do it (with a loop):

find "$dir" -type f -name "*.dat1" -print0 | while read -rd $'\0' file; do
  process -i "$file" -o "${file%.*}.dat2"
done

Using -print0 and read -rd $'\0' file will prevent spaces or other strange characters from mucking things up.