Hi there,
It's pretty hard for me to explain my problem because I'm affraid I'm not using the correct vocabulary. So let me describe the situation. I wrote a script that has one argument. It works like this:

~$ cat /usr/local/bin/squote
echo "$@" | sed 's/'\''/'\''\\'\'\''/g; s/.*/'\''&'\''/g'
~$ squote It\'s great
'It'\''s great'

~$ echo It\'s great | squote
''
~$ cat file | squote
''
~$ squote < file
''

~$ cat /usr/local/bin/squote
cat /dev/stdin | sed 's/'\''/'\''\\'\'\''/g; s/.*/'\''&'\''/g'
echo "$@" | sed 's/'\''/'\''\\'\'\''/g; s/.*/'\''&'\''/g'
~$ echo It\'s great | squote
'It'\''s great'
''
~$ squote It\'s great
# I need to press Ctrl+D
'It'\''s great'

You can test, if the script was called with an argument:

[[ -z "$@" ]] && printf "No input\n" || printf "Input: %q\n" "$@"

Which means: if the length of the arguments ( $@) is zero (-z) then print "No input" else print the string in escaped form. I don't not, what you want to achieve, but if you try to escape strings, check the %q option of printf.

If you call the testscript this way:

testscript "h's m"

it will give you:

Input: h\'s\ m

Thx Christoph,
Although printf is a much better function for my script, you don't answer at all to my problem. What I need is a way to make the script work in the following situation:

~# cat /usr/local/bin/squote
[[ -z "$@" ]] && printf "No input\n" || printf "Input: %q\n" "$@"
~# echo "h's m" | squote
No input