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.rar arg -file ${SRCDIR}/cat01.rar ${SRCDIR}/cat02.rar ${SRCDIR}/cat03.rar... sh run.sh Of course, you can test the run.sh file with different values, or if you have a bunch of files to run the script on, you could do something like this: SRCDIR=`pwd` for i in $FILES do if [ -f "$i" ] then { $GIT checkout origin/$i } SRCDIR=`pwd` COMMAND="${SRCDIR}/scripts/run.sh arg -file "$i" ${SRCDIR}/cat01.rar ${SRCDIR}/cat02.rar ${SRCDIR}/cat03.rar..." exec "$COMMAND" else echo "$i is not a file" fi done Hope it helps. update: It turns out that the command substitution is what is causing the issue. In the above, you are using the result of the git checkout operation (SRCDIR) within the command substitution. Normally this would work, however the action of git checkout is to cache the git repo in the current directory, so the git checkout action is actually creating a git repo at ${SRCDIR}. Therefore, when you do the subsequent command substitution, you are trying to operate on a directory that is not a valid directory. The fix is to use $(SRCDIR) in the command substitution, but that is what you asked for, so now we need to know why you are using command substitution (and not just using a command directly). $COMMAND is the command that was given to sh, not what it actually ran. The command that sh executes is: { $GIT checkout origin/$i } When the shell sees this, it treats this as a string to be passed to the command - Git checkout origin/file.rar. That is where you go wrong - The file $i will be substituted at that point (see Substitutions section of the Shell Programming Guide), and will be substituted with a directory called file. (Note that the dollar sign is not part of the string, but is part of the shell script itself.) Therefore the

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