Partial File Name as Shell Script Input

Question

I am trying to create a shell script that takes a partial file name as its input and then does operations on the files with the matching names. For example, I have three files sample1.txt,sample2.txt and sample3.txt and this is my script

#!bin/bash

VALUE=$1
VALUE2=$2
FILE_NAME=$3

echo $FILE_NAME

And I run it with this command

sh myscript.sh arg1 arg2 sample*

But I get this as the output

sample3.txt (sample2.txt)

But what I want is

sample1.txt
sample2.txt
sample3.txt

How could I do this?

You are passing an argument list from the sender (i.e. sample* could be any number of files). So use an array (list) at the receiver to get all the arguments ("$@") and not just the first — Inian
– Inian, Commented May 21, 2019 at 7:27
do FILE_NAME=("$@") and print the whole array i.e. printf '%s\n' "${FILE_NAME[@]}" — Inian
– Inian, Commented May 21, 2019 at 7:28
@InianWhat if I have other arguments too? I have edited the question — Amber
– Amber, Commented May 21, 2019 at 7:34
Use the shift command to shift the positional arguments. In your case to skip two arguments do shift 2; FILE_NAME=("$@") — Inian
– Inian, Commented May 21, 2019 at 7:39

KamilCuk · Accepted Answer · 2019-05-21 07:51:31Z

1

sample*

get's expanded into (if the files exists, no more files with that name, etc):

sample1.txt sample2.txt sample3.txt

So when you write:

sh myscript.sh arg1 arg2 sample*

What you really write, what your script sees is:

sh myscript.sh arg1 arg2 sample1.txt sample2.txt sample3.txt

Your script get's 5 arguments, not 3

Then you can:

#!bin/bash

VALUE=$1
VALUE2=$2

# shift the arguments to jump over the first two 
shift 2

# print all the rest of arguments
echo "$@"

# ex. read the arguments into an array
FILE_NAME=("$@")
echo "${FILE_NAME[@]}"

Live example at jdoodle.

edited May 21, 2019 at 7:51

answered May 21, 2019 at 7:45

KamilCuk

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