Passing Parameter To Bash Shell Script Through Input Redirection (<)

Question

So, I need to write a shell script that accepts a file path as an argument and uses it within the shell script. I know that I could do something like:

$ ./script filename

With this solution, this would make it so that I could access this filename string using $1 in the shell script.

However, I'm wondering if it is possible to use input redirection instead so I could pass the filename using it:

$ ./script < filename

If so, how would I access this filename in the shell script?
I'm fairly new to shell script coding so I'm not sure if this is even possible.

Answer 1

You can extract the name of file passed with < by listing /proc/<pid>/fd directory, as follows:

ls -ltr /proc/$$/fd

Then

$ cat myscript.sh
ls -ltr /proc/$$/fd
$ ./myscript.sh < hello
total 0
lr-x------ 1 u g 64 Feb 25 08:42 255 -> /tmp/myscript.sh
lrwx------ 1 u g 64 Feb 25 08:42 2 -> /dev/pts/0
lrwx------ 1 u g 64 Feb 25 08:42 1 -> /dev/pts/0
lr-x------ 1 u g 64 Feb 25 08:42 0 -> /tmp/hello

I can't guess whether this is useful

And doesn't work when input is passed through a pipe

$ cat hello| ./myscript.sh 
total 0
lr-x------ 1 u p 64 Feb 25 08:50 255 -> /tmp/myscript.sh
lrwx------ 1 u p 64 Feb 25 08:50 2 -> /dev/pts/0
lrwx------ 1 u p 64 Feb 25 08:50 1 -> /dev/pts/0
lr-x------ 1 u p 64 Feb 25 08:50 0 -> pipe:[82796]

Alternatively, you can use lsof and a little line handling to extract value

 filename=$(/usr/sbin/lsof -p $$| grep " 0u"| cut -c 60-)

Answer 2

You cannot access the filename in the case of ./script < filename but only its contents. The reason is that the shell opens filename and set the corresponding file descriptor to be the standard input of the ./script .

Answer 3

With your second solution you don't access the filename, but your script's standard input is coming from the file.

With the first solution printing the file would look like this:

cat "$1"

With the second:

while read line; do echo "$line"; done