Ubuntu – Why does exclamation mark within double quotes cause a Bash error

bash

Please look at these commands:

$ notify-send SYNC TIME!
$ notify-send 'SYNC TIME!'
$ notify-send "SYNC TIME!"
bash: !": event not found
$

The first two commands produce a notification bubble as expected. The third gives the error shown.

and

$ echo SYNC TIME!
SYNC TIME!
$ echo 'SYNC TIME!'
SYNC TIME!
$ echo "SYNC TIME!"
bash: !": event not found
$

Here as well, the echo works for first two commands but not in the third.

More problems here (although I was not planning on using this): both notify-send "SYNC!TIME" and echo "SYNC!TIME" give bash: !TIME": event not found.

But both notify-send and echo work with "SYNC! TIME"

Can someone please explain why the bash: !": event not found error appears?

Best Answer

! is the default history expansion character in Bash, see the section "HISTORY EXPANSION" in the Bash manpage

History expansion doesn't take place if the ! is enclosed by single quotes, as in
```
notify-send 'SYNC TIME!'
```
History expansion doesn't take place if the ! is followed by a space, tab, newline, carriage return, or =, as in
```
notify-send SYNC TIME!
```
History expansion does take place in
```
echo "SYNC TIME!"
```
So you'll get an error if there isn't a command starting with " in your history

Related Solutions

Ubuntu – Why does the sequence matter in the execution of these bash commands

The time command in your pipeline is not the /usr/bin/time binary, but the bash time built-in. Compare man time with help time. The error you see is bash failing to parse time's argument. This must either be present or be a newline. It is a newline in your first example but absent in the second.

On the other hand, if you were to run

ls;date;'time'

ls;'time';date

where the quotes around 'time' revoke its status as a reserved word, then bash has no problems parsing the line. It now parses three commands in a list, which it will execute in sequence, and /usr/bin/time will report a usage error in either case.

Addendum

It was observed that though time ; date yields an error, time ; ; date does not. The likely explanation is that time ; is interpreted by bash as equivalent to time <newline>. The expression time ; ; date is then parsed as the list of time ; and date.

This is consistent with the observation that time ; and time ; ; are legal as well, the second being parsed as the singleton list containing time ; followed by the optional semicolon allowed after lists.

So another way of explaining why time ; date yields the error bash: syntax error near unexpected token 'date' is that time consumes the semicolon separating it from date. It can only do that because time is a bash reserved word.

Ubuntu – How to run command in an already opened terminal through a shell script

I'd suggested the following as a way to keep the shell open:

gnome-terminal -x sudo -u userA bash -c 'PACKAGE_PATH=/home/userA/package1:/home/userA/package2:$PACKAGE_PATH; \
  cd /home/userA/scripts; \
  source varset.sh; \
  bash'

However, the last bash won't be affected by the assignment of $PACKAGE_PATH or from source varset.sh if they aren't exported.

So:

gnome-terminal -x sudo -u userA bash -c 'export PACKAGE_PATH=/home/userA/package1:/home/userA/package2:$PACKAGE_PATH; \
  cd /home/userA/scripts; \
  source varset.sh; \
  bash'

with varset.sh exported should do the trick.

Best Answer

Related Solutions

Ubuntu – Why does the sequence matter in the execution of these bash commands

Ubuntu – How to run command in an already opened terminal through a shell script

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