I'd like to submit a job to the Sun Grid Engine using a single command line argument, rather than using a shell script. Can it be done?
Example:
I want to do this, and submit to the batch queue:
mkdir empty_directory && cd "empty_directory" && touch "empty_file.txt"
Rather than doing this:
dir_script.sh:
#!/bin/bash
mkdir empty_directory
cd empty_directory
touch empty_file.txt
Then:
qsub dir_script.sh
I do this all the time, using pipe (|):
echo "mkdir empty_directory && cd empty_directory && touch empty_file.txt" | qsub [options]
Related
This question already has answers here:
Why can't I change directories using "cd" in a script?
(33 answers)
Closed 3 years ago.
I'm fairly new to bash scripting.
I'm trying to create a directory with a timestamp and then cd into the directory.
I'm able to create a directory and can cd into it the directory with one command.
mkdir "build" && cd "build"
I can create the directory with data then cd into it.
mkdir date '+%m%d%y' && cd date '+%m%d%y' I am able to create this dir and cd into it with one command.
Here is my bash script:
#!/bin/bash
# mkdir $(date +%F) && cd $(date +%F)
mkdir build_`date '+%m%d%y'` && cd build_`date '+%m%d%y'`
I need to create the directory with build_date '+%m%d%y' in the title then cd into that folder.
I've looked online but am unable to come up with a solution.
Thank you.
Try this:
#!/bin/bash
build_dir="build_$(date '+%m%d%y')"
mkdir $build_dir && cd $build_dir
Best is to name your Folders YYYY.MM.DD like build_2019.11.21
#!/bin/bash
DIRECTORY="build_$(date '+%Y.%m.%d')"
if [ ! -d "$DIRECTORY" ]; then
# checking if $DIRECTORY doesn't exist.
mkdir $DIRECTORY
fi
cd $DIRECTORY
touch testfile
Here's a script that uses /usr/bin/env bash and the builtin command printf instead of /bin/date.
#!/usr/bin/env bash
date="$(printf '%(%m%d%y)T' -1)"
[[ -d build_"${date}" ]] || mkdir build_"${date}"
cd build_"${date}"
# do something useful here
# ...
However, changing directory in a script (subshell) is not much use on its own, as your current working directory ($PWD) will be the same after the script exits.
If you want to have the current shell session change into the new directory, use a shell function.
mkbuild() {
date="$(printf '%(%m%d%y)T' -1)"
[[ -d build_"${date}" ]] || mkdir build_"${date}"
cd build_"${date}"
}
Then run it:
$ echo $PWD
/tmp
$ mkbuild
$ echo $PWD
/tmp/build_112119
I have recently installed Git For Windows version 2.19.1.windows.1 in my Windows 64Bit (both Windows 7 and 10, in two systems).
Now I have created below script to do some regular tasks easily w/o writing every instruction separately:
#!/bin/bash
## Contains functions and methods that can be executed inside Vagrant SSH Session
export SCRIPT_PATH="$(cd "$(dirname "$0")"; pwd -P)"
export PROJECT_ROOT_PATH="/var/www" PROJECT_ROOT_DIR="/var/www" ROOT_FOLDER="/var/www"
function fcc() {
echo "########## Frontend Cache clear begins #############"
[[ "$PWD" =~ "frontend/app" ]] && cd frontend/app
rm -rf webroot/cache_js && rm -rf webroot/cache_css
mkdir -m 777 webroot/cache_js && mkdir -m 777 webroot/cache_css
Console/cake AssetCompress.AssetCompress build -f
cat /dev/null > ~/.bash_history && history -wc && history -cw && exit
cd /var/www
echo "########## Frontend Cache clear ends #############"
keep_shell_open
}
function bocc() {
...
}
function bcc() {
...
}
function succ() {
...
}
function keep_shell_open() { exec $SHELL; };
Now when goto script's directory thro' Gitbash terminal & I register the script as below it unexpectedly opens the C:\Program Files\Git\usr\bin directory of GitBash:
. ./VagrantGuestScript.sh
And then when I execute the "fcc" function of this script, it only executes first 2 or 3 lines and then closes the terminal.
Can anyone explain why it opens Gitbash's bin path when trying to register the script & why it closes the terminal after executing only 2/3 lines of the function "fcc" ?
Let me first write a quick Makefile as a showcase:
#!/bin/make -f
folders := $(shell find -mindepth 1 -maxdepth 1 -type d -print)
make_dir:
#mkdir -p "test0"
pwd_test:
#cd "test0" && pwd
#pwd
pwd_all:
#for f in $(folders); do \
cd "$${f}" && pwd; \
pwd; \
cd ..; \
done
First do make make_dir and then see the different results:
➜ so make pwd_test
/data/cache/tmp/so/test0
/data/cache/tmp/so
➜ so make pwd_all
/data/cache/tmp/so/test0
/data/cache/tmp/so/test0
You see that in the for loop it is necessary to do cd ... Apparently, now there is no child process spawn for the cd X && pwd command, while that is normally the case. Is this behaviour specific to make or specific to my shell?
Make spawns a new process for each command in the rule. Since the for loop is one command you get only one process.
Take a look at Recipe Execution
Edit:
Each line in a makefile gets it own subshell. Commands that have
\ tells make that the next line should be part of the current line.
The reason the for loop get its own subshell is because make see the line as
#for f in $(folders); do cd "$${f}" && pwd; pwd; cd ..; done
MadScientist explains it fairly well. Any command that you can type in your
shell in one line will be executed by make in one subshell or process.
If you were to run this in ksh, ksh would be passed
for f in $(folders); do cd "$${f}" && pwd; pwd; cd ..; done and it would be
run in that one subshell. If ksh did not have a for loop implemented this
probably would error and make would say the command returned some error code.
Explanation of pwd_test
pwd_test:
#cd "test0" && pwd
#pwd
#cd "test0" && pwd is seen as one line so the subshell updates its current
working directory and then prints out what the current working is.
#pwd At this line make spawns a new subshell that contains the old working
directory (or the directory make was called form) and pwd prints that
directory.
I am using capistrano to deploy my php application.I have a requirement to copy a list of configuration files from previous release to new release. The list is maintained in an array. When I am looping over this array to copy from previous to current release , in case the source file is not found it throw an error and further execution stops. I want the script to ignore such case and keep executing next command printing a simple message if source file does not exist. I tried using command like following, but no luck:
run "test -f /tmp/myfile && cp -p /tmp/myfile /home/admin
or even
if(File.exists?("/tmp/myfile"))
run "cp -p /tmp/myfile /home/admin"
else
print " file doesnot exist"
end
Thanks in advance !!
I'd go with this:
run <<-CMD
if [ -f /tmp/myfile ]; then \
cp -p /tmp/myfile /home/admin; \
else \
echo 'myfile does not exist'; \
fi
CMD
Remember that all capistrano run commands are executed on the remote server, and
only an exit value of 0 indicates success.
The result of "test -f /tmp/myfile && cp -p /tmp/myfile /home/admin would
still be 1 if /tmp/myfile does not exist. You could use || to
call echo with a message that the file does not exist:
test -f /tmp/myfile && cp -p /tmp/myfile /home/admin && echo myfile does not exist
Which is the same thing.
I've looked around for an answer to this one but couldn't find one.
I have written a simple script that does initial server settings and I'd like it to remove/unlink itself from the root directory on completion. I've tried a number of solutions i googled ( for example /bin/rm $test.sh) but the script always seems to remain in place. Is this possible? Below is my script so far.
#! /bin/bash
cd /root/
wget -r -nH -np --cut-dirs=1 http://myhost.com/install/scripts/
rm -f index.html* *.gif */index.html* */*.gif robots.txt
ls -al /root/
if [ -d /usr/local/psa ]
then
echo plesk > /root/bin/INST_SERVER_TYPE.txt
chmod 775 /root/bin/*
/root/bin/setting_server_ve.sh
rm -rf /root/etc | rm -rf /root/bin | rm -rf /root/log | rm -rf /root/old
sed -i "75s/false/true/" /etc/permissions/jail.conf
exit 1;
elif [ -d /var/webmin ]
then
echo webmin > /root/bin/INST_SERVER_TYPE.txt
chmod 775 /root/bin/*
/root/bin/setting_server_ve.sh
rm -rf /root/etc | rm -rf /root/bin | rm -rf /root/log | rm -rf /root/old
sed -i "67s/false/true/" /etc/permissions/jail.conf
break
exit 1;
else
echo no-gui > /root/bin/INST_SERVER_TYPE.txt
chmod 775 /root/bin/*
/root/bin/setting_server_ve.sh
rm -rf /root/etc | rm -rf /root/bin | rm -rf /root/log | rm -rf /root/old
sed -i "67s/false/true/" /etc/permissions/jail.conf
break
exit 1;
fi
rm -- "$0"
Ought to do the trick. $0 is a magic variable for the full path of the executed script.
This works for me:
#!/bin/sh
rm test.sh
Maybe you didn't really mean to have the '$' in '$test.sh'?
The script can delete itself via the shred command (as a secure deletion) when it exits.
#!/bin/bash
currentscript="$0"
# Function that is called when the script exits:
function finish {
echo "Securely shredding ${currentscript}"; shred -u ${currentscript};
}
# Do your bashing here...
# When your script is finished, exit with a call to the function, "finish":
trap finish EXIT
The simplest one:
#!/path/to/rm
Usage: ./path/to/the/script/above
Note: /path/to/rm must not have blank characters at all.
I wrote a small script that adds a grace period to a self deleting script based on
user742030's answer https://stackoverflow.com/a/34303677/10772577.
function selfShred {
SHREDDING_GRACE_SECONDS=${SHREDDING_GRACE_SECONDS:-5}
if (( $SHREDDING_GRACE_SECONDS > 0 )); then
echo -e "Shreding ${0} in $SHREDDING_GRACE_SECONDS seconds \e[1;31mCTRL-C TO KEEP FILE\e[0m"
BOMB="●"
FUZE='~'
SPARK="\e[1;31m*\e[0m"
SLEEP_LEFT=$SHREDDING_GRACE_SECONDS
while (( $SLEEP_LEFT > 0 )); do
LINE="$BOMB"
for (( j=0; j < $SLEEP_LEFT - 1; j++ )); do
LINE+="$FUZE"
done
LINE+="$SPARK"
echo -en $LINE "\r"
sleep 1
(( SLEEP_LEFT-- ))
done
fi
shred -u "${0}"
}
trap selfShred EXIT
See the repo here: https://github.com/reedHam/self-shred
$0 may not contain the script's name/path in certain circumstances. Please check the following: https://stackoverflow.com/a/35006505/5113030 (Choosing between $0 and BASH_SOURCE...)
The following script should work as expected in these cases:
source script.sh - the script is sourced;
./script.sh - executed interactively;
/bin/bash -- script.sh - passed as an argument to a shell program.
#!/usr/bin/env bash
# ...
rm -- "$( readlink -f -- "${BASH_SOURCE[0]:-$0}" 2> '/dev/null'; )";
Please check the following regarding shell script source reading and execution since it may affect the behavior when a script is deleted while running: https://unix.stackexchange.com/a/121025/133353 (How Does Linux deal with shell scripts?...)
Related: https://stackoverflow.com/a/246128/5113030 (How can I get the source directory of a Bash script from...)
Just add to the end:
rm -- "$0"
Why remove the script at all? As other have mentioned it means you have to keep a copy elsewhere.
A suggestion is to use a "firstboot" like approach. Simply create an empty file in e.g. /etc/sysconfig that triggers the execution of this script if it is present. Then remove that file at the end of the script.
Modify the script so it has the necessary chkconfig headers and place it in /etc/init.d/ so it is run at every boot.
That way you can rerun the script at a later time simply by recreating the trigger script.
Hope this helps.