Shell Programming and Scripting

Hi Gurus

I have to write a script which does something like this

/usr/local/gpg2/bin/GPG2 -o arch_696354351_1_23.dbf -d arch_696354351_1_23.dbf.gpg
/usr/local/gpg2/bin/GPG2 -o arch_696354351_1_24.dbf -d arch_696354351_1_24.dbf.gpg
/usr/local/gpg2/bin/GPG2 -o arch_696354351_1_25.dbf -d arch_696354351_1_25.dbf.gpg
/usr/local/gpg2/bin/GPG2 -o arch_696354351_1_26.dbf -d arch_696354351_1_26.dbf.gpg
/usr/local/gpg2/bin/GPG2 -o arch_696354351_1_27.dbf -d arch_696354351_1_27.dbf.gpg

all my arch files would be in /ARCH directory.

I'm not sure what's your question nor your problem, I guess maybe you're looking for something like this?

Code:

$ for ((i=23; i<=27; i++)); do echo /usr/local/gpg2/bin/GPG2 -o arch_696354351_1_$i.dbf -d arch_696354351_1_$i.dbf.gpg; done
/usr/local/gpg2/bin/GPG2 -o arch_696354351_1_23.dbf -d arch_696354351_1_23.dbf.gpg
/usr/local/gpg2/bin/GPG2 -o arch_696354351_1_24.dbf -d arch_696354351_1_24.dbf.gpg
/usr/local/gpg2/bin/GPG2 -o arch_696354351_1_25.dbf -d arch_696354351_1_25.dbf.gpg
/usr/local/gpg2/bin/GPG2 -o arch_696354351_1_26.dbf -d arch_696354351_1_26.dbf.gpg
/usr/local/gpg2/bin/GPG2 -o arch_696354351_1_27.dbf -d arch_696354351_1_27.dbf.gpg

i am not sure how will this work.

i could do something like

for i in `ls -l ./*.dbf | grep -v "^gpg"' `
do
/usr/local/gpg2/bin/GPG2 -o $i -d *.gpg
done


but when i run this i get this error
bash: command substitution: line 1: unexpected EOF while looking for matching `''
bash: command substitution: line 2: syntax error: unexpected end of file

the problem is instead of putting -d *.gpg in the GPG2 command i need to have the complete file name including .gpg for example arch_696354351_1_23.dbf.gpg

so -o in the command would take arch_696354351_1_23.dbf and -d would take arch_696354351_1_23.dbf.gpg

Code:

for i in *.dbf; do /usr/local/gpg2/bin/GPG2 -o $i -d $i.gpg; done

Code:

bash-2.05# for i in *.dbf
do
/usr/local/gpg2/bin/GPG2 -o $i -d $i.gpg
done

usage: gpg [options] --decrypt [filename]

it need the file name correctly

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My guess:

Code:

for i in $( ls *.dbf$ )