pass outfile to another program using bash script - linux

Question

I have the following bash code for a batch run of multiple files to be processed by 3 different programs:
for i in *.txt
do
program1 -in ${i} -out Prog1_out_${i}
program2 -in Prog1_out_${i} -out Prog2_out_${i}
program3 -in Prog2_out_${i} -out Prog3_out_${i}
done
I ran into problem with program 2 not finding the input which is the output from program 1, and of course program 3 did not find the required input.
Can anyone help with suggestions for solving the problem?
Thanks

Answer 1

If the programs produce the output whenever they are successful, you could make them dependent of the previous commands success, like this:
program1 -in ${i} -out Prog1_out_${i} &&
program2 -in Prog1_out_${i} -out Prog2_out_${i} &&
program3 -in Prog2_out_${i} -out Prog3_out_${i}
So if one of the programs fails, the rest of the chain will not be invoked.
However, if the creation of the output has nothing to do with the success of the program, but you just want to check if the files exist, you can add the appropriate check before you call programx, i.e.
if [ -f "${i}" ]
then
progx ...
fi
As you are doing the same thing all the time this could be generalized for all programs (untested):
for i in *.txt
do
mv $i Prog0_out_$i
for program in 0 1 2
do
INFILE=Prog{$program}_out_${$i}
if [ ! -r ${INFILE} ]
then
break
fi
program{$program} -in ${INFILE} -out "Prog{$program}_out_$((i+1))"
done
done