BASH SCRIPT of LS command

Shell Programming and Scripting
1

I need help in writing a BASH SCRIPT of ls command.

for example:
$ ./do_ls.sh files
f1.txt
f2.jpeg
f3.doc

$ ./do_ls.sh dirs
folder1
folder2
folder3

My attempt:

#!/bin/bash
#
if test $# -d file
then
	echo $dirs
	else
if test $# -f file
then
	echo $files
	exit 1

thanks.

2

try something like this..

#!/bin/bash
#
var=$1
if [[ "$var" == "files" ]]
then
ls -l| sed -n '/^-/p'
elif [[ "$var" == "dirs" ]]
then
ls -l| sed -n '/^d/p'
else
echo "wrong choice"
fi
3 
#!/bin/bash
if [[ $1 == "files" ]]
then
 ls -p|sed -n '/\/$/!p'
elif [[ $1 == "dirs" ]]
then
 ls -p|sed -n 's:/$::p'
fi
4

Thanks for the reply.

I found this in my old book as a subsitue for ls command.

#!/bin/bash
#
for f in * ; do echo $f ; done

But here, I want to display only files when $ ./do_ls.sh files is entered.

and dir when $ ./do_ls.sh dirs is entered.

---------- Post updated at 02:23 PM ---------- Previous update was at 12:17 PM ----------

bump..