Bash Script argument disappears

wahming · 10-25-2010, 05:20 AM

I call my script with ./script -e arg2
However, $@ only shows me arg2. Surprisingly, if I use ./script -s arg2, $@ shows me both arguments. What's going on?

GrapefruiTgirl · 10-25-2010, 05:31 AM

I don't see an invocation option of -e in the bash manpage, and -s doesn't seem as though it would do as you describe..

So, figuring that this has something to do with the code in the script, it would perhaps help to see the code. What do the -e and -s options do within the code? Anything?

Also, you're only passing one arg in your examples, so how can either example show you both args?

Kenhelm · 10-25-2010, 06:28 AM

-e is an option to echo but -s is not.

Code:

echo -e arg2
arg2

echo -s arg2
-s arg2

This is one way to get -e to echo, but it creates a leading space in the output.

Code:

echo '' -e arg2
 -e arg2

When echoing to the screen the leading space can be removed using $' \b', which in bash is a space followed by a backspace.

Code:

echo $' \b'-e arg2
-e arg2

wahming · 10-25-2010, 09:34 PM

Grapefruit - sorry, I was a little sleepy when I posted, so maybe my question wasn't as clear as it might have been. -e would be arg1, and arg2 would be, well, arg2. `echo $@` in my script would only display arg2, but not arg1 in some cases.

Kenhelm - thanks! You hit the nail on the head, and printf "%s " works beautifully.

GrapefruiTgirl · 10-26-2010, 05:03 AM

No problem! Kenhelm did indeed hit it right on.

If that solves the issue fully, you can mark this thread solved if you like, using Thread Tools menu, above the first post.

Have a good day!