Testing presence of multiple files within loop

Question

Assume a bash script to which four commandline arguments are passed. Each of these arguments represents a path to, and a name of, an input file (i.e., INF1, INF2, INF3, INF4). I would like to evaluate for each of these four input files, if it exists and - if a file does not exist as specified - exit the script.

#!/bin/bash

# Assigning commandline arguments
INF1=$1
INF2=$2
INF3=$3
INF4=$4

# Check if input files exist
if [ ! -f $1 ]; then
    echo -ne " ERROR | File not found: $1\n"
    exit 1
fi
if [ ! -f $2 ]; then
    echo -ne " ERROR | File not found: $2\n"
    exit 1
fi
if [ ! -f $3 ]; then
    echo -ne " ERROR | File not found: $3\n"
    exit 1
fi
if [ ! -f $4 ]; then
    echo -ne " ERROR | File not found: $4\n"
    exit 1
fi

I would think that having four individual if-statements here is unnecessary, and that the same functionality can be achieved with a single if-statement, wrapped into a loop. How can I reduce the code in this aspect?

karakfa · Accepted Answer · 2018-04-03 18:01:42Z

2

in a script

#!/bin/bash

for v; do
if [ ! -f "$v" ]; then
    echo "ERROR | File not found: $v"
    exit 1
fi
done

edited Apr 3, 2018 at 18:01

answered Apr 3, 2018 at 17:04

karakfa

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1 Comment

Michael Over a year ago

You can even skip the in "$@" part, because that's the default if you don't have an in clause.

umran akaalp · Accepted Answer · 2018-04-03 17:00:49Z

1

You can declare an array and assing the variables and loop through it.

ARRAY=()
ARRAY+=$1
ARRAY+=$2 ..

and

for i in "${arrayName[@]}"
do
   : 
   # do whatever on $i
done

answered Apr 3, 2018 at 17:00

umran akaalp

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