I am messing around with shell scripts and I am trying to get my script to take input from another command, like say ls. It is called like this:
ls | ./example.sh
I try to access the input from within example.sh
#!/bin/bash
echo $1
but it echoes back nothing. Is there another way to reference parameters given to bash by other commands, because it works if I type in:
./example.sh poo
Parameters and input aren't the same thing:
$ ls
example.sh foo
$
$ cat example.sh
for param; do
printf 'argument 1: "%s"\n' "$param"
done
while IFS= read -t 1 -r input; do
printf 'input line: "%s"\n' "$input"
done
$
$ ls | ./example.sh
input line: "example.sh"
input line: "foo"
$
$ ls | ./example.sh bar
argument 1: "bar"
input line: "example.sh"
input line: "foo"
$
$ ./example.sh $(ls)
argument 1: "example.sh"
argument 1: "foo"
Parameters are the arguments passed to the script to start it running, input is what the script reads while it's running.
You can use xargs to do that.
ls | xargs ./example.sh
Resource
https://ss64.com/bash/xargs.html
https://www.cyberciti.biz/faq/linux-unix-bsd-xargs-construct-argument-lists-utility/
Or read man page for it
