#!/bin/bash
for arg
do
echo "$arg"
done
In file called compile.
I tried this:
compile ha -aa -bb -cc -dd -ee -ff -gg
ha
-aa
-bb
-cc
-dd
-ff
-gg
Why does -ee not show up? In fact, it seems that -[e]+ does not show up.
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2 Answers
Because the version of echo you use takes -e as an option (meaning "expand escape characters").
Edited to add:
The standard response to this type of question is "use printf", which is strictly accurate but somewhat unsatisfactory. printf is a lot more annoying to use because of the way it handles multiple arguments. Thus:
$ echo -e a b c
a b c
$ printf "%s\n" -e a b c
-e
a
b
c
$ printf "%s" -e a b c
-eabc$ # That's not what I wanted either
Of course, you just need to remember to quote the entire argument sequence, but that can lead to annoying quote-escaping issues.
Consequently, I offer for your echoing pleasure:
$ ech-o() { printf "%s\n" "$*"; }
$ ech-o -e a b c
-e a b c
$
Comments
Here it says that it's not possible with GNU echo but you could use printf.
for arg
do
printf "%s\n" "$arg"
done
