Create a shell script to execute multiple commands

Question

Here is what i want to do... I want to write a script which sends all the files in a folder as an input to the command2.

I know pipes are used to send the output of one command to another but in my case, when i do

ls | command

What this does is it sends the output of ls ( i.e. all the files in the folder ) as a single input. Instead, I want that the individual files are sent as an argument to the command2, one by one.

Can anyone please help me out i really searched a lot for it but didn't find much.

https://stackoverflow.com/questions/3886295/how-do-i-list-one-filename-per-output-line-in-linux would that help? — Sonamor, Apr 09 '18 at 16:29
Chances are that `find` or a loop of files in your directory will do a better job of whatever you are trying to do. It's [generally considered a bad idea to parse the output](https://unix.stackexchange.com/questions/128985/why-not-parse-ls) of `ls` — JNevill, Apr 09 '18 at 16:30
In this particular case, of course, simply `command *` is superior, and avoids [parsing `ls` output](https://mywiki.wooledge.org/ParsingLs) — tripleee, Apr 09 '18 at 17:02
@Sonamor : I understood the question in that the OP want to invoke *command* many times, once for each input file. — user1934428, Apr 10 '18 at 05:44
@ChintanMehta: How about `echo *|xargs -n 1 your_command` ? Note that in this case, *your_command* doesn't get the (single) filename via stdin, but as command line argument. — user1934428, Apr 10 '18 at 05:46

score 2 · Answer 1 · answered Apr 09 '18 at 16:30

2

use any of the below methods, but i recommend the first one, since no subshell is been created

for files in *;do
   if [[ -f "$files" ]];then
       echo "$files"
   fi
done


while read line;do
   if [[ -f "$line" ]];then
       echo "$line"
   fi
done < <(ls)

answered Apr 09 '18 at 16:30

0.sh

2,659
16
37

test -f is useful to avoid a directory – kyodev Apr 09 '18 at 17:24
I want to encrypt those files with gpg command. Now your first script runs well, but when i replace the 3rd line with, _gpg -e -r "Recipient"_ I get the following output _usage: gpg [options] --encrypt [filename]_ Instead of running that command – Chintan Mehta Apr 09 '18 at 17:24
the fault is not from the script, the fault is from your `gpg` command, what is "Recipient" ? – 0.sh Apr 09 '18 at 17:27
recipient is the user whose key it is. BTW, i have used the same gpg command without the shellscript and it works flawlessly so i don't think its that @0.sh – Chintan Mehta Apr 09 '18 at 17:29
bro pass in the filename to gpg `gpg -e -r "Recipient" "$files"` – 0.sh Apr 09 '18 at 17:29
problem solved ?? @ChintanMehta – 0.sh Apr 09 '18 at 17:36
Yeah!! used _gpg -e -r "Chintan Mehta" $eachfilename_ Thanks!! – Chintan Mehta Apr 09 '18 at 17:41
@ChintanMehta accept the answer – 0.sh Apr 09 '18 at 17:49

score 0 · Answer 2 · answered Apr 09 '18 at 16:32

0

You can use find, for example:

$ cat do.sh
#!/usr/bin/env sh

printf "arg: %s\n" "$1"
$ touch {1..10}
$ find . -type f -exec ./do.sh {} \;
arg: ./6
arg: ./10
arg: ./5
arg: ./3
arg: ./9
arg: ./2
arg: ./8
arg: ./do.sh
arg: ./1
arg: ./4
arg: ./7

answered Apr 09 '18 at 16:32

Arkadiusz Drabczyk

11,227
2
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I didn't understand what you're doing here. And yes, I tried running this file in my folder. It didn't work – Chintan Mehta Apr 09 '18 at 17:23
What file? What didn't work? – Arkadiusz Drabczyk Apr 09 '18 at 17:24
Nevermind, I used another method and it worked. Do check out in the answers. – Chintan Mehta Apr 09 '18 at 17:39

score 0 · Answer 3 · answered Apr 09 '18 at 16:54

You can use xargs.

$ find . -maxdepth 1 -print0 | xargs -0 -l command

this will print all of the files in your current directory, separating the names by null chars, then xargs take them, one at a time (-l) and invokes command using each file as an argument.

Create a shell script to execute multiple commands

3 Answers3