In a script, I've defined: rm() { echo "rm $#" }
Normally rm would run echo rm, but I want to change it for a single line to run the actual rm command. How can I do this?
I've tried:
rm= rm file but this still runs echo rm file
I don't want to unset rm because I still want to be able to run rm file and expect it to echo.
Note: rm is just an example command, not the command I'm actually using.
You can use command to run a command instead of a function by the same name:
#!/bin/bash
hostname() {
echo "You are in a maze of twisty little passages, all alike. "
}
hostname # runs the function
command hostname # runs the external command
Related
I am writing a ksh to checkout the code and setup the compilation variables through xenv setup. This is how my script looks at the moment -
#!/usr/bin/ksh
logname=$LOGNAME
homedir="$HOME/${logname}-SVN-Dev/pkgroot"
#Create directory <username>-SVN-Dev to contain copied code.
if [ -z "$logname" ]
then
logname=`/usr/ucb/whoami`
fi
RunCmd "rm -rf $homedir"
RunCmd "mkdir -p $homedir"
## Some code to checkout code
cd $HOME
echo "setenv PKGROOT $homedir">>$HOME/.cshrc
echo "setenv DEVROOT $homedir/src">>$HOME/.cshrc
source $HOME/.cshrc
RunCmd "/xenv/xenv -L -i $homedir/My.env $homedir;"
make -f project.mk createmakefile
The xenv above switches the shell and goes into a new prompt, ia_cross: and my shell exits without executing the "make -f project.mk createmakefile" command.
I have tries putting a pipe between make and xenv but that didn't solve this. Any suggestions would really help?
use below command in your script,
echo "make -f project.mk createmakefile" | /xenv/xenv -c -i $homedir/My.env $homedir;
This question already has answers here:
Why can't I change directories using "cd" in a script?
(33 answers)
Closed 7 years ago.
What I'm trying to do
I've created a shell script that I've added to my $PATH that will download and get everything setup for a new Laravel project. I would like the script to end by changing my terminal directory into the new project folder.
From what I understand right now currently it's only changing the directory of the sub shell where the script is actually running. I can't seem to figure out how to do this. Any help is appreciated. Thank you!
#! /usr/bin/env bash
echo -e '\033[1;30m=========================================='
## check for a directory
if test -z "$1"; then
echo -e ' \033[0;31m✖ Please provide a directory name'
exit
fi
## check if directory already exist
if [ ! -d $1 ]; then
mkdir $1
else
echo -e ' \033[0;31m✖ The '"$1"' directory already exists'
exit
fi
# move to directory
cd $1
## Download Laravel
echo -e ' \033[0;32m+ \033[0mDownloading Laravel...'
curl -s -L https://github.com/laravel/laravel/zipball/master > laravel.zip
## Unzip, move, and clean up Laravel
echo -e ' \033[0;32m+ \033[0mUnzipping and cleaning up files...'
unzip -q laravel.zip
rm laravel.zip
cd *-laravel-*
mv * ..
cd ..
rm -R *-laravel-*
## Make the /storage directory writable
echo -e ' \033[0;32m+ \033[0mMaking /storage directory writable...'
chmod -R o+w storage
## Download and install the Generators
echo -e ' \033[0;32m+ \033[0mInstalling Generators...'
curl -s -L https://raw.github.com/JeffreyWay/Laravel-Generator/master/generate.php > application/tasks/generate.php
## Update the application key
echo -e ' \033[0;32m+ \033[0mUpdating Application Key...'
MD5=`date +”%N” | md5`
sed -ie 's/YourSecretKeyGoesHere!/'"$MD5"'/' application/config/application.php
rm application/config/application.phpe
## Create .gitignore and initial git if -git is passed
if [ "$2" == "-git" ]; then
echo -e ' \033[0;32m+ \033[0mInitiating git...'
touch .gitignore
curl -s -L https://raw.github.com/gist/4223565/be9f8e85f74a92c95e615ad1649c8d773e908036/.gitignore > .gitignore
# Create a local git repo
git init --quiet
git add * .gitignore
git commit -m 'Initial commit.' --quiet
fi
echo -e '\033[1;30m=========================================='
echo -e ' \033[0;32m✔ Laravel Setup Complete\033[0m'
## Change parent shell directory to new directory
## Currently it's only changing in the sub shell
filepath=`pwd`
cd "$filepath"
You can technically source your script to run it in your parent shell instead of spawning a subshell to run it. This way whatever changes you make to your current shell (including changing directories) persist.
source /path/to/my/script/script
or
. /path/to/my/script/script
But sourcing has its own dangers, use carefully.
(Peripherally related: how to use scripts to change directories)
Use a shell function to front-end your script
setup () {
# first, call your big script.
# (It could be open-coded here but that might be a bit ugly.)
# then finally...
cd someplace
}
Put the shell function in a shell startup file.
Child processes (including shells) cannot change current directory of parent process. Typical solution is using eval in the parent shell. In shell script echo commands you want to run by parent shell:
echo "cd $filepath"
In parent shell, you can kick the shell script with eval:
eval `sh foo.sh`
Note that all standard output will be executed as shell commands. Messages should output to standard error:
echo "Some messages" >&2
command ... >&2
This can't be done. Use exec to open a new shell in the appropriate directory, replacing the script interpreter.
exec bash
I suppose one possibility would be to make sure that the only output of your script is the path name you want to end up in, and then do:
cd `/path/to/my/script`
There's no way your script can directly affect the environment (including it's current directory) of its parent shell, but this would request that the parent shell itself change directories based on the output of the script...
this topic has been discussed at length, however, I have a variant on the theme that I just cannot crack. Two days into this now and decided to ping the community. THx in advance for reading..
Exec. summary is I have a script in OS X that runs fine and executes without issue or error when done manually. When I put the script in the crontab to run daily it still runs but it doesnt run all of the commands (specifically SFTP).
I have read enough posts to go down the path of environment issues, so as you will see below, I hard referenced the location of the SFTP in the event of a PATH issue...
The only thing that I can think of is the IdentityFile. NOTE: I am putting this in the crontab for my user not root. So I understand that it should pickup on the id_dsa.pub that I have created (and that has already been shared with the server)..
I am not trying to do any funky expect commands to bypass the password, etc. I dont know why when run from the cron that it is skipping the SFTP line.
please see the code below.. and help is greatly appreciated.. thx
#!/bin/bash
export DATE=`date +%y%m%d%H%M%S`
export YYMMDD=`date +%y%m%d`
PDATE=$DATE
YDATE=$YYMMDD
PATH=/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin
FEED="~/Dropbox/"
USER="user"
HOST="host.domain.tld"
A="/tmp/5nPR45bH"
>${A}.file1${PDATE}
>${A}.file2${PDATE}
BYEbye ()
{
rm ${A}.file1${PDATE}
rm ${A}.file2${PDATE}
echo "Finished cleaning internal logs"
exit 0
}
echo "get -r *" >> ${A}.file1${PDATE}
echo "quit" >> ${A}.file1${PDATE}
eval mkdir ${FEED}${YDATE}
eval cd ${FEED}${YDATE}
eval /usr/bin/sftp -b ${A}.file1${PDATE} ${USER}#${HOST}
BYEbye
exit 0
Not an answer, just comments about your code.
The way to handle filenames with spaces is to quote the variable: "$var" -- eval is not the way to go. Get into the habit of quoting all variables unless you specifically want to use the side effects of not quoting.
you don't need to export your variables unless there's a command you call that expects to see them in the environment.
you don't need to call date twice because the YYMMDD value is a substring of the DATE: YYMMDD="${DATE:0:6}"
just a preference: I use $HOME over ~ in a script.
you never use the "file2" temp file -- why do you create it?
since your sftp batch file is pretty simple, you don't really need a file for it:
printf "%s\n" "get -r *" "quit" | sftp -b - "$USER#$HOST"
Here's a rewrite, shortened considerably:
#!/bin/bash
PATH=/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin
FEED_DIR="$HOME/Dropbox/$(date +%Y%m%d)"
USER="user"
HOST="host.domain.tld"
mkdir "$FEED_DIR" || { echo "could not mkdir $FEED_DIR"; exit 1; }
cd "$FEED_DIR"
{
echo "get -r *"
echo quit
} |
sftp -b - "${USER}#${HOST}"
I'm trying to write a bash script that "wraps" whatever the user wants to invoke (and its parameters) sourcing a fixed file just before actually invoking it.
To clarify: I have a "ConfigureMyEnvironment.bash" script that must be sourced before starting certain executables, so I'd like to have a "LaunchInMyEnvironment.bash" script that you can use as in:
LaunchInMyEnvironment <whatever_executable_i_want_to_wrap> arg0 arg1 arg2
I tried the following LaunchInMyEnvironment.bash:
#!/usr/bin/bash
launchee="$#"
if [ -e ConfigureMyEnvironment.bash ];
then source ConfigureMyEnvironment.bash;
fi
exec "$launchee"
where I have to use the "launchee" variable to save the $# var because after executing source, $# becomes empty.
Anyway, this doesn't work and fails as follows:
myhost $ LaunchInMyEnvironment my_executable -h
myhost $ /home/me/LaunchInMyEnvironment.bash: line 7: /home/bin/my_executable -h: No such file or directory
myhost $ /home/me/LaunchInMyEnvironment.bash: line 7: exec: /home/bin/my_executable -h: cannot execute: No such file or directory
That is, it seems like the "-h" parameter is being seen as part of the executable filename and not as a parameter... But it doesn't really make sense to me.
I tried also to use $* instead of $#, but with no better outcoume.
What I'm doing wrong?
Andrea.
Have you tried to remove double quotes in exec command?
Try this:
#!/usr/bin/bash
typeset -a launchee
launchee=("$#")
if [ -e ConfigureMyEnvironment.bash ];
then source ConfigureMyEnvironment.bash;
fi
exec "${launchee[#]}"
That will use arrays for storing arguments, so it will handle even calls like "space delimited string" and "string with ; inside"
Upd: simple example
test_array() { abc=("$#"); for x in "${abc[#]}"; do echo ">>$x<<"; done; }
test_array "abc def" ghi
should give
>>abc def<<
>>ghi<<
You might want to try this (untested):
#!/usr/bin/bash
launchee="$1"
shift
if [ -e ConfigureMyEnvironment.bash ];
then source ConfigureMyEnvironment.bash;
fi
exec "$launchee" $#
The syntax for exec is exec command [arguments], however becuase you've quoted $launchee, this is treated as a single argument - i.e., the command, rather than a command and it's arguments. Another variation may be to simply do: exec $#
Just execute it normally without exec
#!/usr/bin/bash
launchee="$#"
if [ -e ConfigureMyEnvironment.bash ];
then source ConfigureMyEnvironment.bash;
fi
$launchee
Try dividing your list of argumets:
ALL_ARG="${#}"
Executable="${1}"
Rest_of_Args=${ALL_ARG##$Executable}
And try then:
$Executable $Rest_of_Args
(or exec $Executable $Rest_of_Args)
Debugger
I have a script with a number of options in it one of the option sets is supposed to change the directory and then exit the script however running over ssh with the source to get it to change in the parent it exits SSH is there another way to do this so that it does not exit? my script is in the /usr/sbin directory.
You might try having the script run a subshell instead of whatever method it is using to “change [the directory] in the parent” (presumably you have the child print out a cd command and have the parent do something like eval "$(script --print-cd)"). So instead of (e.g.) a --print-cd option, add a --subshell option that starts a new instance of $SHELL.
d=/path/to/some/dir
#...
cd "$d"
#...
if test -n "$opt_print_cd"; then
sq_d="$(printf %s "$d" | sed -e "s/'/'\\\\''/g")"
printf "cd '%s'\n" "$sq_d"
elif test -n "$opt_subshell"; then
exec "$SHELL"
fi
If you can not edit the script itself, you can make a wrapper (assuming you have permission to create new, persistent files on the ‘server’):
#!/bin/sh
script='/path/to/script'
print_cd=
for a; do test "$a" = --print-cd && print_cd=yes && break; done
if test -n "$print_cd"; then
eval "$("$script" ${1+"$#"})" # use cd instead of eval if the script prints a bare dir path
exec "$SHELL"
else
exec $script" ${1+"$#"}
fi