How to iterate over positional parameters in a Bash script?

Question

Where am I going wrong?

I have some files as follows:

filename_tau.txt
filename_xhpl.txt
fiename_fft.txt
filename_PMB_MPI.txt
filename_mpi_tile_io.txt

I pass tau, xhpl, fft, mpi_tile_io and PMB_MPI as positional parameters to script as follows:

./script.sh tau xhpl mpi_tile_io fft PMB_MPI

I want grep to search inside a loop, first searching tau, xhpl and so on..

point=$1     #initially points to first parameter
i="0"
while [$i -le 4]
do
  grep "$str" ${filename}${point}.txt
  i=$[$i+1]
  point=$i     #increment count to point to next positional parameter
done

Answer 1

See here, you need shift to step through positional parameters.

Answer 2

Set up your for loop like this. With this syntax, the loop iterates over the positional parameters, assigning each one to 'point' in turn.

for point; do
  grep "$str" ${filename}${point}.txt 
done

Answer 3

Try something like this:

# Iterating through the provided arguments
for ARG in $*; do
    if [ -f filename_$ARG.txt]; then
        grep "$str" filename_$ARG.txt 
    fi
done

Answer 4

args=$@;args=${args// /,}
grep "foo" $(eval echo file{$args})

Answer 5

There is more than one way to do this and, while I would use shift, here's another for variety. It uses Bash's indirection feature:

#!/bin/bash
for ((i=1; i<=$#; i++))
do
    grep "$str" ${filename}${!i}.txt
done

One advantage to this method is that you could start and stop your loop anywhere. Assuming you've validated the range, you could do something like:

for ((i=2; i<=$# - 1; i++))

Also, if you want the last param: ${!#}