I have a bash function i would like to bsub. It is recursively called when i try to source the script, but if i don't source the script it doesn't seem to recognize my function. How do I correctly call besub on a function within the same script file?
my_script(should print "12345"):
#! /bin/sh
function myFunct {
echo $1
}
bsub -q myQueue "source ./my_script; myFunct 12345"
a.bash may look like this
#! /bin/bash
export input=$1
function myFunct {
echo "$input"
}
# This is if you want to call bsub outside the bash
# Use bsub -q Queue `./a.bash 12345`
myFunct "$input"
# Put bsub inside and call the script
# ./a.bash 12345
bsub -q myQueue `myFunct "$input"`
I've gotten it to work with
bsub [...] bash -rcfile my_script.sh -i -c "myFunct 12345"
Related
I need to create a file for subsequent nohup execution, the file contains some variables, I don't want the variables to be replaced when I generate through the file.
When I executed the code with the following code, all the variables were replaced in the generated script, which was not what I expected.
#!/bin/bash
gen_script() {
filepath=$1
if [ ! -f "$filepath" ]; then
cat >${filepath} <<EOF
#!/bin/bash
# Code generated by main.sh; DO NOT EDIT.
test(){
ip=$1
port=$2
restorefile=$3
redis-cli -h $ip -p $port --pipe < $restorefile
}
test "$#"
EOF
fi
}
main(){
gen_script exec.sh
nohup bash exec.sh $1 $2 > nohup.out 2>&1 &
}
main "$#"
How can I change my code please? I really appreciate any help with this.
To disable expansions in here document, quote the delimieter:
cat <<'EOF'
... $not_expanded
EOF
Instead, let bash serialize the function.
#!/bin/bash
work() {
ip=$1
port=$2
restorefile=$3
redis-cli -h $ip -p $port --pipe < $restorefile
}
gen_script() {
filepath=$1
if [ ! -f "$filepath" ]; then
cat >${filepath} <<EOF
#!/bin/bash
# Code generated by main.sh; DO NOT EDIT.
$(declare -f work)
work "\$#"
EOF
fi
}
main() {
gen_script exec.sh
nohup bash exec.sh "$1" "$2" > nohup.out 2>&1 &
}
main "$#"
Check your script with shellcheck. Do not define a function named test, there is already a super standard command test which is an alias for command [.
I have a bash script named test.sh with:
#!/bin/bash
var1=hello
sh test2.sh $var1="/var/log"
My test2.sh looks like this:
#!/bin/bash
function jumpto
{
label=$1
cmd=$(sed -n "/$label:/{:a;n;p;ba};" $0 | grep -v ':$')
eval "$cmd"
exit
}
start=${1:-"start"}
jumpto $start
start:
echo "variable -- " $var1
This does not work due to I used jumpto function. When execution, always $var1 assigned to the $1 variable in jumpto function.
Is there a different way to do this.?
We suggest to learn about environment variable and export command and their scope.
Try modify your test.sh
#!/bin/bash
export var1=hello
sh test2.sh $var1="/var/log"
I've been working through creating a script to move some files from a local machine to a remote server. As part of that process I have a function that can either be called directly or wrapped with 'declare -fp' and sent along to an ssh command. The code I have so far looks like this:
export REMOTE_HOST=myserver
export TMP=eyerep-files
doTest()
{
echo "Test moving files from $TMP with arg $1"
declare -A files=(["abc"]="123" ["xyz"]="789")
echo "Files: ${!files[#]}"
for key in "${!files[#]}"
do
echo "$key => ${files[$key]}"
done
}
moveTest()
{
echo "attempting move with wrapped function"
ssh -t "$REMOTE_HOST" "$(declare -fp doTest|envsubst); doTest ${1#Q}"
}
moveTest $2
If I run the script with something like
./myscript.sh test dev
I get the output
attempting move with wrapped function
Test moving files from eyerep-files with arg dev
Files: abc xyz
bash: line 7: => ${files[]}: bad substitution
It seems like the string expansion for the for loop is not working correctly. Is this expected behaviour? If so, is there an alternative way to loop through an array that would avoid this issue?
If you're confident that your remote account's default shell is bash, this might look like:
moveTest() {
ssh -t "$REMOTE_HOST" "$(declare -f doTest; declare -p $(compgen -e)); doTest ${1#Q}"
}
If you aren't, it might instead be:
moveTest() {
ssh -t "$REMOTE_HOST" 'exec bash -s' <<EOF
set -- ${##Q}
$(declare -f doTest; declare -p $(compgen -e))
doTest \"\$#\"
EOF
}
I managed to find an answer here: https://unix.stackexchange.com/questions/294378/replacing-only-specific-variables-with-envsubst/294400
Since I'm exporting the global variables, I can get a list of them using compgen and use that list with envsubst to specify which variables I want to replace. My finished function ended up looking like:
moveTest()
{
echo "attempting move with wrapped function"
ssh -t "$REMOTE_HOST" "$(declare -fp doTest|envsubst "$(compgen -e | awk '$0="${"$0"}"') '${1}'"); doTest ${1#Q}"
}
Say I have a bash script and I want some variables to appear when sourced and others to only be accessible from within the script (both functions and variables). What's the convention to achieve this?
Let's say test.sh is your bash script.
What you can do is extract all the common items and put them in common.sh which can be sourced by other scripts.
The BASH_SOURCE array helps you here:
Consider this script, source.sh
#!/bin/bash
if [[ ${BASH_SOURCE[0]} == "$0" ]]; then
# this code is run when the script is _executed_
foo=bar
privFunc() { echo "running as a script"; }
main() {
privFunc
publicFunc
}
fi
# this code is run when script is executed or sourced
answer=42
publicFunc() { echo "Hello, world!"; }
echo "$0 - ${BASH_SOURCE[0]}"
[[ ${BASH_SOURCE[0]} == "$0" ]] && main
Running it:
$ bash source.sh
source.sh - source.sh
running as a script
Hello, world!
Sourcing it:
$ source source.sh
bash - source.sh
$ declare -p answer
declare -- answer="42"
$ declare -p foo
bash: declare: foo: not found
$ publicFunc
Hello, world!
$ privFunc
bash: privFunc: command not found
$ main
bash: main: command not found
How to get the command previously used to start a shell script?
for example:
nohup /script_name.sh &
Inside the script itself, how to check if "nohup" has been used?
Thanks.
You want to use the $_ parameter in your script.
Example: shell.sh
#!/bin/bash
echo $_;
user#server [~]# sh shell.sh
/usr/bin/sh
user#server [~]#
Additionally:
If you want to get rid of that full path - /usr/bin/sh - utilize basename command.
#!/bin/bash
echo `basename $_`;
user#server [~]# sh shell.sh
sh
user#server [~]#
well that depends on the script in question.There're many ways to execute a script like:
./<scriptname> #chmod 700 <scriptname> should be done before executing this script
bash <scriptname> # provided bash is used for executing the script.
or if you just want to get the name of script2 in script1, then use sed or awk for parsing the script1 with regular expression => /script2/.
Try this:
cat <script1> | awk '{ if( $0 ~ /^[^#].* \/scriptname.sh/ ){ print $1}}'
#codebaus thanks, doing something like this works but using strace definitely not.
#!/bin/bash
# echo $_
# echo $0
if grep "sh" $_ >/dev/null ; then
exit 1
fi ;
echo "string" ;
I believe you want to run this?:
#!/bin/bash
# echo $_
# echo $0
if grep "sh" $_ 2> /dev/null ; then
exit 1
fi ;
echo "string";
user#server [~]# sh shell.sh
Binary file /usr/bin/sh matches
user#server [~]#
Not sure what you are trying to accomplish in the end game. But $_ should give you what you need based on your initial question.
Additionally:
As I did not answer your strace comment, apologies. Based on the previous code above.
strace sh shell.sh
wait4(-1, Binary file /usr/bin/strace matches
[{WIFEXITED(s) && WEXITSTATUS(s) == 0}], 0, NULL) = 874
rt_sigprocmask(SIG_SETMASK, [], NULL, 8) = 0
--- SIGCHLD (Child exited) # 0 (0) ---