Bash printf %q invalid directive

问题:

I want to change my PS1 in my .bashrc file. I've found a script using printf with %q directive to escape characters :

#!/bin/bash
STR=$(printf "%q" "PS1=\u@\h:\w\$ ")
sed -i '/PS1/c\'"$STR" ~/.bashrc

The problem is that I get this error :

script.sh: 2: printf: %q: invalid directive

Any idea ? Maybe an other way to escape the characters ?

回答1:

The printf command is built into bash. It's also an external command, typically installed in /usr/bin/printf. On most Linux systems, /usr/bin/printf is the GNU coreutils implementation.

Older releases of the GNU coreutils printf command do not support the %q format specifier; it was introduced in version 8.25, released 2016-10-20. bash's built-in printf command does -- and has as long as bash has had a built-in printf command.

The error message implies that you're running script.sh using something other than bash.

Since the #!/bin/bash line appears to be correct, you're probably doing one of the following:

sh script.sh
. script.sh
source script.sh

Instead, just execute it directly (after making sure it has execute permission, using chmod +x if needed):

./script.sh

Or you could just edit your .bashrc file manually. The script, if executed correctly, will add this line to your .bashrc:

PS1=\\u@\\h:\\w\$\ 

(The space at the end of that line is significant.) Or you can do it more simply like this:

PS1='\u@\h:\w\$ '

One problem with the script is that it will replace every line that mentions PS1. If you just set it once and otherwise don't refer to it, that's fine, but if you have something like:

if [ ... ] ; then
    PS1=this
else
    PS1=that
fi

then the script will thoroughly mess that up. It's just a bit too clever.

回答2:

Keith Thompson has given good advice in his answer. But FWIW, you can force bash to use a builtin command by preceding the command name with builtin eg

builtin printf "%q" "PS1=\u@\h:\w\$ "

Conversely,

command printf "%s\n" some stuff

forces bash to use the external command (if it can find one).

command can be used to invoke commands on disk when a function with the same name exists. However, command does not invoke a command on disk in lieu of a Bash built-in with the same name, it only works to suppress invocation of a shell function. (Thanks to Rockallite for bringing this error to my attention).

It's possible to enable or disable specific bash builtins (maybe your .bashrc is doing that to printf). See help enable for details. And I guess I should mention that you can use

type printf

to find out what kind of entity (shell function, builtin, or external command) bash will run when you give it a naked printf. You can get a list of all commands with a given name by passing type the -a option, eg

type -a printf 

---

You can use grep to see the lines in your .bashrc file that contain PS1:

grep 'PS1' ~/.bashrc 

or

grep -n0 --color=auto 'PS1=' ~/.bashrc

which gives you line numbers and fancy coloured output. And then you can use the line number to force sed to just modify the line you want changed.

Eg, if grep tells you that the line you want to change is line 7, you can do

sed -i '7c\'"$STR" ~/.bashrc

to edit it. Or even better,

sed -i~ '7c\'"$STR" ~/.bashrc

which backs up the original version of the file in case you make a mistake.

When using sed -i I generally do a test run first without the -i so that the output goes to the shell, to let me see what the modifications do before I write them to the file.