I have configured a Jenkins job to source a bash script that sources another bash script which adds an alias to the .bashrc of its user and sources the .bashrc itself, and then original script tries to use that alias (set by the second). However, it cannot seem to find the alias it has just created. I am not using any scripting plugins aside from using a "Send files or execute commands over SSH" build step to source the script.
The job does this:
source ./test_script.sh
test_script.sh looks like this:
echo "In test_script.sh"
echo $USER
echo $HOME
source ./setup_env.sh
echo "\nBack in test_script.sh"
alias foo
foo
And finally, setup_env.sh looks like this:
echo "\nIn setup_env.sh"
echo "alias foo=\"echo foobar\"" >> $HOME/.bashrc
source $HOME/.bashrc 2>/dev/null
cat $HOME/.bashrc
The output I receive from the Jenkins job looks like this:
In test_script.sh
my_user
/home/my_user
\nIn setup_env.sh
...all of my bashrc...
alias foo="echo foo"
\nBack in test_script.sh
alias foo='echo foo'
./test_script.sh: line 7: foo: command not found
I don't understand why this is happening, when I can happily run it myself on the command-line and watch it succeed. Why can't Jenkins use the new alias, when it can obviously find it (as demonstrated by the output of the alias foo command)?
For anyone else who's having this problem, you may need to set the expand_aliases shell option, which seems to be off by default with Jenkins:
shopt expand_aliases # check if it's on
shopt -s expand_aliases # set expand_aliases option to true
shopt expand_aliases # it should be on now
# test with a simple alias, should print 1
alias x="python -c 'print 1'"
x
The \n that is showing in the output of your echo commands
suggest this is not running under bash, as you may be expecting.
Please check the setting of your jenkins user
(or any other user that you run Jenkins with) -
especially the setting of the default shell.
To test this, add one of those lines at the beginning of your script:
env
or
env | grep -i shell
Should also consider making sure your scripts run under the correct shell,
by adding the "shebang" line as the first line in each script.
In the case of 'bash', for example, you should add the following first line:
#!/bin/bash
(it is not a comment, despite what the auto-syntax-highlighter may think...)
Related
Is there a way to tell a bash script not to import any variables from the parent shell i.e. ignore exported variables. There is such capability in slurm --export=NONE so I wonder if there is an option I can put in the #!/bin/bash line to get similar behavior.
On linux the cleanest option I found was:
#!/usr/bin/env -S - bash
env
which for me prints:
PWD=/home/allan
SHLVL=1
_=/usr/bin/env
Another option is:
#!/usr/bin/env bash
[ -n "$HOME" ] && exec -c "$0"
env
Possible using $BASH_SOURCE[0] instead of $0 as the latter can be set by user. $BASH_SOURCE, however, is not always set. Hard-coding the script path would work but that's ugly.
I know I can run an "original" command (not alias) using either \ or "":
\ls
"ls"
This doesn't work for functions though. Also it requires me to use that syntax every time.
Is it possible in a sourced script to disable all functions/aliases from the parent process (one which runs my script)? I.e. if a user in their terminal has some aliases functions defined I want them disabled in my script (but of course I still want to be able to define and use aliases/functions of my own).
Types of Commands in Bash
Bash knows different types of commands which can shadow each other. The precedence of these types is:
aliases
can be defined by the user using alias cmd=...
functions
can be defined by the user using cmd() { ... }
built-ins
are directly implement in bash and cannot be altered. help and enable list all built-ins.
Executable files in $PATH
Meaning if you type cmd arg1 arg2 ... you use the alias cmd if it is defined, otherwise you use the function cmd if it is defined, otherwise you use the built-in cmd if it is built-in, otherwise you use the first executable cmd from the directories in $PATH if there is one, otherwise you end up with the error -bash: cmd command not found.
Which of these cases applies for cmd can be checked using type -a cmd.
Manual Precedence Control
Bash allows you to influence which type to pick using quoting and the built-ins command and builtin.
\cmd
suppresses aliases
uses functions, built-ins, executables
command cmd
suppresses aliases and functions
uses built-ins and executables
builtin cmd
supresses aliases, functions, and executables
uses only built-ins
enable -n cmd
disables the built-in cmd completely, such that afterwards only
aliases, functions, and executables are used
env cmd
not a bash built-in, therefore it doesn't really suppress anything but
uses only executables
Examples
Shadowing is perfectly normal. For instance, bash has its own built-in echo, but your system also has /bin/echo. Both implementations may differ. For instance, my echo from bash 5 supports \uXXXX but my echo from GNU coreutils 8.3 does not. The possibility of such differences becomes even more clear if you add your own implementations using aliases and functions. Here's an example in an interactive bash session ($ is the prompt):
$ echo() { printf "function echo: %s\n" "$*"; }
$ alias echo='printf "alias echo: %s %s %s\n"'
$ type -a echo
echo is aliased to `printf "alias echo: %s %s %s\n"'
echo is a function
echo ()
{
printf "function echo: %s\n" "$*"
}
echo is a shell builtin
echo is /bin/echo
$ echo -e '\u2261'
alias echo: -e \u2261
$ \echo -e '\u2261'
function echo: -e \u2261
# use the built-in (or executable file if there was no such built-in)
$ command echo -e '\u2261'
≡
$ builtin echo -e '\u2261'
≡
# use the executable /bin/echo
$ env echo -e '\u2261'
\u2261
$ enable -n echo
# use the executable /bin/echo (`command` is needed to skip the alias and function)
$ command echo -e '\u2261'
\u2261
Answering your Question
Unfortunately I'm not aware of something like enable to permanently disable alias and function lookup. You could try some hacks like backing up all aliases and functions, doing unset -f and unalias on them, and restoring them at the end. However, unset may fail for readonly functions. The better way would be to use bash -c '... functions and aliases have no effect here ...' for the parts where you don't really need the benefits of source. For the other parts, prefix everything with command.
Please note: The caller who sources your script may even disable or shadow command, builtin, and so on -- therefore you can never be sure that you are actually using the commands you expected. Even writing /usr/bin/env executable or /path/to/the/executable does not help as a function can have the name and $PATH or the file system can be altered.
However, that shouldn't be your concern. The one who sources your script should be responsible for providing the correct environment.
Edit: this answer might no longer be relevant since you edited the question to clarify that the script is being sourced, not being executed in a subshell.
This happens by default. Proof:
$ function x() { echo 'hi'; }
$ x
hi
$ bash
# We are now in a subshell.
$ x
bash: x: command not found
Functions are often defined in one of the shell's startup files: .bashrc, .profile or .bash_profile. Which of these are sourced depends on whether the shell is a login shell and/or an interactive shell. A shell that invoked to execute a shell script is neither a login shell nor an interactive shell, and in this case none of those files are sourced.
EDIT: I should read more carefully, as you don't want to source a script, but be sourced, the following is for the other way around:
Functions
If you source your parent script at the beginning, you can just loop through the defined functions and unset them.
declare -F will list all defined functions but in the format declare -f functioname, so you have to get only the name:
IFS=$'\n'
for f in $(declare -F|cut -d ' ' -f 3); do
unset -f $f
done
Aliases
Alias should not be sourced in as i remember, but if they are there you can do
unalias -a
to unset them all.
Could you help me, why this script works when sourced (or even directly on console) and does not work on a script?
I have checked and in any case I'm using the same bash in /bin/ and always 4.4.19(1)-release (checked with $BASH_VERSION).
Moreover I tried removing shebang but nothing changes.
#!/bin/bash
fname=c8_m81l_55.fit
bname=${fname%%+(_)+([0-9]).fit}
echo $bname
GIving these results:
test:~$ ./test.sh
c8_m81l_55.fit
test:~$ . ./test.sh
c8_m81l
Bash does not recognize +(pattern) syntax unless extglobs are enabled, and they are disabled by default. Apparently your bash setup enables them in interactive sessions; that's why your script works only when sourced in an interactive shell.
To fix that, either enable extglobs within the script by this command:
shopt -s extglob
Or use an alternative that works irrespective of shell's interactiveness:
bname=$(sed 's/__*[0-9][0-9]*\.fit$//' <<< $fname)
# with GNU sed it'd look like:
bname=$(sed -E 's/_+[0-9]+\.fit$//' <<< $fname)
$0 expands to the name of the shell script.
$ cat ./sample-script
#!/bin/bash
echo $0
$ chmod 700 ./sample-script
$ ./sample-script
./sample-script
If the shell script is invoked via a symbolic link, $0 expands to its name:
$ ln -s ./sample-script symlinked-script
$ ./symlinked-script
./symlinked-script
How could I get the name of an alias? Here `$0' expands again to the filename:
$ alias aliased-script=./sample-script
$ aliased-script
./sample-script
Aliases are pretty dumb, according to the man page
...Aliases are expanded when a command is read, not when it is executed...
so since bash is basically just replacing a string with another string and then executing it, there's no way for the command to know what was expanded in the alias.
I imagine you already know this, but for the record the answer is: you need cooperation by the code implementing the alias.
alternate_name () {
MY_ALIAS_WAS=alternate_name real_name "$#"
}
or, if you really want to use the superseded alias syntax:
alias alternate_name="MY_ALIAS_WAS=alternate_name real_name"
...and then...
$ cat ~/bin/real_name
#!/bin/sh
echo $0, I was $MY_ALIAS_WAS, "$#"
bash does not make this available. This is why symlinks are used to invoke multiplex commands, and not aliases.
This question already has answers here:
Why aliases in a non-interactive Bash shell do not work
(4 answers)
Closed 6 years ago.
I am trying to execute a command remotely over ssh, example:
ssh <user>#<host> <command>
The command which needs to be executed is an alias, which is defined in .bashrc, e.g.
alias ll='ls -al'
So what in the end the following command should get executed:
ssh user#host "ll"
I already found out that .bashrc only gets sourced with interactive shell, so in .bash_login I put:
if [ -f ~/.bashrc ]; then
. ~/.bashrc
fi
and I also tried to define the alias directly in .bash_login.
I also tried to put the alias definition / sourcing of .bashrc in .bash_profile and also in .ssh/rc. But nothing of this works.
Note that I am not able to change how the ssh command is invoked since this is a part of some binary installation script. The only thing I can modify is the environment. Is there any other possibility to get this alias sourced when the ssh command is executed? Is there some ssh configuration which has to be adapted?
From the man pages of bash:
Aliases are not expanded when the shell is not interactive, unless the expand_aliases shell option is set using shopt
There are a couple ways to do this, but the simplest is to just add the following line to your .bashrc file:
shopt -s expand_aliases
Instead of:
ssh user#host "bash -c ll"
try:
ssh user#host "bash -ic ll"
to force bash to use an "interactive shell".
EDIT:
As pointed out here about non-interactive shells..
# If not running interactively, don't do anything
[ -z "$PS1" ] && return
# execution returns after this line
Now, for every alias in your bashrc file say i have:
alias ll="ls -l"
alias cls="clear;ls"
Create a file named after that alias say for ll:
user#host$ vi ssh_aliases/ll
#inside ll,write
ls -l
user#host$ chmod a+x ll
Now edit .bashrc to include:
# If not running interactively, don't do anything
[ -z "$PS1" ] && export $PATH=$PATH:~/ssh_aliases
This does the job.. although I am not sure if it is the best way to do so
EDIT(2)
You only need to do this for aliases, other commands in bashrc will be executed as pointed out by David "you must have executable for ssh to run commands".
an alternative to alias that will be visible in all script is
EXPORT & EXECUTE VARIABLE
# shortcut to set enviroment to insensitive case
export go_I="shopt -s nocasematch"
Now in any script you can use
#!/bin/bash
$go_I # go Insensitive
[[ a == A ]] # evaluates TRUE ( $? == 0)
$go_C # maibe want to go back to casesensitive
it's useful to place all shortcuts/aliases in /path/to/my_commands and edit /etc/bash.bashrc
source /path/to/my_commands
Open file ~/.bash_profile. If this file does not exist create one in the home directory and add the below line
source = $HOME/.bashrc
exit your ssh and login agian and you should get the .bashrc settings working for you.