I have a script I'm trying to make at least somewhat platform-independent. Because I can't count on the PATHs or shell aliases of my users (some of whom have been known to create aliases matching "reserved words" meaningful to the shell), I've taken to aliasing all non-builtin shell commands with uppercase equivalents. This works fine when I do it semi-manually, e.g.
AWK=/usr/bin/awk
DATE=/bin/date
GREP=/bin/grep
, but not so well when I try it in a function by iterating through an array of commands:
createAliases() {
COMMANDS=(awk chmod date echo git mkdir)
WHICH=/usr/bin/which
for command in ${COMMANDS[#]}; do
${command^^}=$($WHICH ${command})
done
}
, which produces errors like AWK=/usr/bin/awk: No such file or directory. Is my attempt to use the string^^ up-casing mechanism interfering with the variable assignment? It looks like the shell is actually trying to run the (aliased) commands, which is not what I want.
Any assistance is appreciated!
The following seems to work:
shopt -s expand_aliases
createAliases() {
COMMANDS=(awk chmod date echo git mkdir)
WHICH=/usr/bin/which
for command in ${COMMANDS[#]}; do
alias ${command^^}=$($WHICH ${command})
done
}
Prefixing the assigment with alias actually registers the desired aliases.
The line shopt -s expand_aliases enables you to then use these alias from anywhere in the script, according to https://www.thegeekdiary.com/how-to-make-alias-command-work-in-bash-script-or-bashrc-file/
Related
I am trying to use the solution of using sudo on my existing aliases as covered in many existing answers already and i am so confused as to why it is not working
alias sudo='sudo '
I keep all my aliases in .bash_aliases. Inside .bash_aliases I have this
function _test {
echo 'test!'
}
alias test='_test'
I reload the .bashrc each time; source .bashrc but when I run sudo test I always get
sudo: _test: command not found
The only strange thing that happens is that I get the following on reload and not on a new terminal
dircolors: /home/MYHOME/.dircolors: No such file or directory
but i feel this is a red herring.
As l0b0 says, aliases cannot be used in this way in bash.
However, you can pass a function through (and really, there's basically never a good reason to use an alias instead of sticking to functions alone).
_test() {
echo 'test!'
}
sudo_test() {
sudo bash -c "$(declare -f _test)"'; _test "$#"' sudo_test "$#"
}
...will define a command sudo_test that runs the function _test via sudo. (If your real-world version of this function calls other functions or requires access to shell variables, add those other functions to the declare -f command line, and/or add a declare -p inside the same command substitution to generate a textual description of variables your function needs).
To run an alias like alias_name it must be exactly the first word in the command, so sudo alias_name will never work. Ditto 'alias_name', \alias_name and other things which eventually expand to the alias name.
I am trying to use the solution of using sudo on my existing aliases as covered in many existing answers already and i am so confused as to why it is not working
alias sudo='sudo '
I keep all my aliases in .bash_aliases. Inside .bash_aliases I have this
function _test {
echo 'test!'
}
alias test='_test'
I reload the .bashrc each time; source .bashrc but when I run sudo test I always get
sudo: _test: command not found
The only strange thing that happens is that I get the following on reload and not on a new terminal
dircolors: /home/MYHOME/.dircolors: No such file or directory
but i feel this is a red herring.
As l0b0 says, aliases cannot be used in this way in bash.
However, you can pass a function through (and really, there's basically never a good reason to use an alias instead of sticking to functions alone).
_test() {
echo 'test!'
}
sudo_test() {
sudo bash -c "$(declare -f _test)"'; _test "$#"' sudo_test "$#"
}
...will define a command sudo_test that runs the function _test via sudo. (If your real-world version of this function calls other functions or requires access to shell variables, add those other functions to the declare -f command line, and/or add a declare -p inside the same command substitution to generate a textual description of variables your function needs).
To run an alias like alias_name it must be exactly the first word in the command, so sudo alias_name will never work. Ditto 'alias_name', \alias_name and other things which eventually expand to the alias name.
I wrote simple command that lets me run the last N commands from terminal history. It looks like this: $> r 3 which will replay the last 3 commands.
I have the following alias in my bash profile:
alias r="history -w; runlast $1"
And then the following simple perl script for the runlast command:
#!/usr/bin/env perl
use strict;
use warnings;
my $lines = $ARGV[0] || exit;
my #last_commands = split /\n/,
`bash -ic 'set -o history; history' | tail -$lines`;
#last_commands =
grep { $_ !~ /(^r |^history |^rm )/ }
map { local $_ = $_; s/^\s+\d+\s+//; $_ }
#last_commands;
foreach my $cmd (#last_commands) {
system("$cmd");
}
This works but my bash profile has aliases and other features (e.g. color output) I want the perl script to have access to. How do I load the bash profile for perl so it runs the bash commands with my bash profile? I read somewhere that if you "source the bash profile" for perl you can get it to work. So I tried adding source ~/.bash_profile; to my r command alias but that didn't have an effect. I'm not sure if I was doing that correctly, though.
The system forks a process in which it runs a shell, which is non-login and non-interactive; so no initialization is done and you get no aliases. Also note that the shell used is /bin/sh, which is generally a link to another shell. This is often bash but not always, so run bash explicitly.
To circumvent this you need to source the file with aliases, but as bash man page says
Aliases are not expanded when the shell is not interactive, unless the expand_aliases shell option is set using shopt (see the description of shopt under SHELL BUILTIN COMMANDS below).
Thus you need shopt -s expand_aliases, as mentioned. But there is another screw: on that same physical line aliases are not yet available; so it won't work like this in a one-liner.
I'd also recommend to put aliases in .bashrc, or in a separate file that is sourced.
Solutions
Add shopt -s expand_aliases to your ~/.bashrc, and before the aliases are defined (or the file with them sourced), and run bash as a login shell
system('/bin/bash', '-cl', 'source ~/.bashrc; command');
where -l is short for --login.
In my tests the source ~/.bashrc wasn't needed; however, the man page says
When bash is invoked as an interactive login shell, or as a non-interactive shell with the --login option, it first reads and executes commands from the file /etc/profile, if that file exists. After reading that file, it looks for ~/.bash_profile, ~/.bash_login, and ~/.profile, in that order, and reads and executes commands from the first one that exists and is readable.
and goes on to specify that ~/.bashrc is read when an interactive shel that is not login runs. So I added explicit sourcing.
In my tests sourcing .bashrc (with shopt added) while not running as a login shell didn't work, and I am not sure why.
This is a little heavy-handed. Also, initialization may be undesirable to run from a script.
Source ~/.bashrc and issue shopt command, and then a newline before the command
system('/bin/bash', '-c',
'source ~/.bashrc; shopt -s expand_aliases\ncommand');
Really. It works.
Finally, is this necessary? It asks for trouble, and there is probably a better design.
Other comments
The backticks (qx) is context-aware. If it's used in list context – its return assigned to an array, for example – then the command's output is returned as a list of lines. When you use it as the argument for split then it is in the scalar context though, when all output is returned in one string. Just drop split
my #last_commands = `bash -ic 'set -o history; history $lines`;
where I also use history N to get last N lines. In this case the newlines stay.
history N returns last N lines of history so there is no need to pipe to last
Regex substitution in a map can be done without changing the original
map { s/^\s+\d+\s+//r } #last_commands;
With /r modifier the s/// operator returns the new string, not changing the original. This "non-destructive substitution" has been available since v5.14
No need to explicitly use $_ in the last grep, and no need for parenthesis in regex
grep { not /^r |^history |^rm ?/ } ...
or
grep { not /^(?:r|history|rm)[ ]?/ } ...
where parens are now needed, but as it is only for grouping the ?: makes it not capture the match. I use [ ] to emphasize that that space is intended; this is not necessary.
I also added ? to make space optional since history (and r?) may have no space.
The proper solution is to have your Perl script just print the commands, and make your current interactive shell eval the string printed from your history. (I would probably get rid of Perl entirely but that's beside the point here.)
If the commands get evaluated in the current shell, you avoid many contextual problems which would be very hard or even intractable with system() or generally anything involving a new process. For example, a subprocess cannot have access to non-exported variables in the current shell. var="foo", echo "$var"; r 1 is going to be very hard to solve correctly with your current approach. Using the current interactive shell will also naturally and easily solve the problems you were having with trying to get a noninteractive subshell act like an interactive one.
Aliases suck anyway, so let's redefine r as a function:
r(){
history -w
eval $(printlast "$1")
}
... where refactoring runlast into a different script printlast is a trivial additional requirement. Or maybe just turn it into a (much simpler!) shell function:
printlast () {
history "$1" |
perl -ne 's/^\s*\d+\s+\*?//; print unless m/^(history|rm?)($|\s)'
}
With this, you can also get rid of history -w from the r definition.
Notice how we are using Perl where it is useful; but the main functionality makes sense to keep in the shell when you're dealing with the shell.
You can't source in a Bash script into a Perl script. The bash_profile has to be sourced in by the shell that executes the command. When Perl runs system, it forks a new shell each time.
You have to source in the bash_profile for each command that you run through system:
system('source ~/.bash_profile; ' + $cmd);
One more thing, system invokes a non-interactive shell. So, your Bash aliases defined in .bash_profile won't work unless you invoke:
shopt -s expand_aliases
inside that script
This question already has answers here:
How to use aliases defined in .bashrc in other scripts?
(6 answers)
Closed 2 years ago.
I'm using a Mac and I have this alias defined in .bashrc:
$cat .bashrc | grep la
alias la='ls -la'
then I try to use it in a script:
$cat ./mytest.sh
#!/bin/bash
la
It runs and says it cannot find la:
./mytest.sh: line 2: la: command not found
Why is this? I tried on both Mac and Linux, same error!
Your .bashrc is only used by interactive shells. https://www.gnu.org/software/bash/manual/bashref.html#Bash-Startup-Files says:
Invoked non-interactively
When Bash is started non-interactively, to run a shell script, for example, it looks for the variable BASH_ENV in the environment, expands its value if it appears there, and uses the expanded value as the name of a file to read and execute. Bash behaves as if the following command were executed:
if [ -n "$BASH_ENV" ]; then . "$BASH_ENV"; fi
but the value of the PATH variable is not used to search for the filename.
As noted above, if a non-interactive shell is invoked with the --login option, Bash attempts to read and execute commands from the login shell startup files.
As you can see, there's nothing about .bashrc there. Your alias simply doesn't exist in the script.
But even if .bashrc were read, there's another problem:
Aliases are not expanded when the shell is not interactive, unless the expand_aliases shell option is set using shopt.
So if you wanted aliases to work in a script, you'd have to do shopt -s expand_aliases first. Or just use a shell function instead of an alias:
la() {
ls -la
}
The simplest answer is to fix this issue is to do the 2 important things in your script -or it wont' work, if you just do one thing.
#!/bin/bash -i
# Expand aliases defined in the shell ~/.bashrc
shopt -s expand_aliases
After this, your aliases that you have defined in ~/.bashrc they will be available in your shell script (giga.sh or any.sh) and to any function or child shell within such script.
If you don't do that, you'll get an error:
your_cool_alias: command not found
At the beginning of the ~/.bashrc file usually can be found two lines as:
# If not running interactively, don't do anything
[ -z "$PS1" ] && return
This line aborts the inclusion for scripts which is not recommended anyway. For portability issues, you usually write the full command or define the alias in the script itself.
This question already has answers here:
How to use aliases defined in .bashrc in other scripts?
(6 answers)
Closed 2 years ago.
I'm using a Mac and I have this alias defined in .bashrc:
$cat .bashrc | grep la
alias la='ls -la'
then I try to use it in a script:
$cat ./mytest.sh
#!/bin/bash
la
It runs and says it cannot find la:
./mytest.sh: line 2: la: command not found
Why is this? I tried on both Mac and Linux, same error!
Your .bashrc is only used by interactive shells. https://www.gnu.org/software/bash/manual/bashref.html#Bash-Startup-Files says:
Invoked non-interactively
When Bash is started non-interactively, to run a shell script, for example, it looks for the variable BASH_ENV in the environment, expands its value if it appears there, and uses the expanded value as the name of a file to read and execute. Bash behaves as if the following command were executed:
if [ -n "$BASH_ENV" ]; then . "$BASH_ENV"; fi
but the value of the PATH variable is not used to search for the filename.
As noted above, if a non-interactive shell is invoked with the --login option, Bash attempts to read and execute commands from the login shell startup files.
As you can see, there's nothing about .bashrc there. Your alias simply doesn't exist in the script.
But even if .bashrc were read, there's another problem:
Aliases are not expanded when the shell is not interactive, unless the expand_aliases shell option is set using shopt.
So if you wanted aliases to work in a script, you'd have to do shopt -s expand_aliases first. Or just use a shell function instead of an alias:
la() {
ls -la
}
The simplest answer is to fix this issue is to do the 2 important things in your script -or it wont' work, if you just do one thing.
#!/bin/bash -i
# Expand aliases defined in the shell ~/.bashrc
shopt -s expand_aliases
After this, your aliases that you have defined in ~/.bashrc they will be available in your shell script (giga.sh or any.sh) and to any function or child shell within such script.
If you don't do that, you'll get an error:
your_cool_alias: command not found
At the beginning of the ~/.bashrc file usually can be found two lines as:
# If not running interactively, don't do anything
[ -z "$PS1" ] && return
This line aborts the inclusion for scripts which is not recommended anyway. For portability issues, you usually write the full command or define the alias in the script itself.