Bash – using ! character in echo

bashechospecial characters

I am messing with echo and having issues with echo displaying the ! character.

I tried this.

echo -e "Wake!\nUp!
bash: !\n!: event not found

Then this

# echo "Wake\!\nUp!"
Wake\!
Up!

Well now the backslash AND the exclamation mark shows. How can I use this character properly? What am I missing?

Best Answer

Use single quotes:

$ echo -e 'Wake!\nUp!'
Wake!
Up!

If you use single quotes ('') shell will treat the string literally, whereas if you use double quotes ("") it will treat ! as a reference to the previous command (event).

Related Solutions

Echo Bang – How to Echo a Bang!

Try using single quotes.

echo -e '#!/bin/bash \n /usr/bin/command args'  > .scripts/command

echo '#!'

echo '#!/bin/bash'

The problem is occurring because bash is searching its history for !/bin/bash. Using single quotes escapes this behaviour.

Bash – Properly Escape Exclamation Points in Bash

Use single quotes:

echo -e '#!/usr/bin/python\nimport string as s,random;print "".join(random.sample(s.letters+s.digits+s.punctuation,9))'>pg;chmod +x pg;./pg

The rules for ! were sort of grafted onto the other quoting rules afterwards (from csh). They were very useful back when shells didn't have command line editing, but some people still use them now.

P.S. Since you're coding for bash:

echo $'#!/usr/bin/python\nimport string as s,random;print"".join(random.sample(s.letters+s.digits+s.punctuation,9))'>pg;chmod +x pg;./pg

This works on most unices:

echo python -c \''import string as s,random;print"".join(random.sample(s.letters+s.digits+s.punctuation,9))'\'>pg;chmod +x pg;./pg

(Not that I understand why you want to create a script or why the script name has to be two letters.)

Best Answer

Related Solutions

Echo Bang – How to Echo a Bang!

Bash – Properly Escape Exclamation Points in Bash

Related Question