I have two shell scripts for two different user user 'a' and user 'root':
a.sh for user 'a'
export CODE=`pwd | cut -d / -f 5`
isEnabled=`grep -i isEnabled $HOME/$CODE/config/code.properties | awk -F'=' '{print $2}'`
b.sh for user 'root'
su - a -c "sh /export/home/a/a1/bin/a.sh status
As you can see, root execute script b.sh which execute script a.sh
but actually it failed since it does not find the path to the script since the
variable $CODE in a.sh probably missing in run time while running b.sh
after investigation i tired the following in b.sh:
. /path/to/a.sh
source /path/to/a.sh
. /path/to/a.sh "$CODE"
. $(/path/to/a.sh $CODE)
Can someone please advise?
Thank you all.
The variable $CODE should be there, but it will be set to a's home directory. Try using basename $0 instead of pwd. That will get the directory the script is being executed from, rather than the current working directory of user executing it.
Related
a.sh
#! /bin/sh
export x=/usr/local
we can do source ./a in command-line. But I need to do the export through shell script.
b.sh
#! /bin/sh
. ~/a.sh
no error... but $x in command-line will show nothing. So it didn't get export.
Any idea how to make it work?
a.sh
#! /bin/sh
export x=/usr/local
-----------
admin#client: ./a.sh
admin#client: echo $x
admin#client: <insert ....>
You can put export statements in a shell script and then use the 'source' command to execute it in the current process:
source a.sh
You can't do an export through a shell script, because a shell script runs in a child shell process, and only children of the child shell would inherit the export.
The reason for using source is to have the current shell execute the commands
It's very common to place export commands in a file such as .bashrc which a bash will source on startup (or similar files for other shells)
Another idea is that you could create a shell script which generates an export command as it's output:
shell$ cat > script.sh
#!/bin/sh
echo export foo=bar
^D
chmod u+x script.sh
And then have the current shell execute that output
shell$ `./script.sh`
shell$ echo $foo
bar
shell$ /bin/sh
$ echo $foo
bar
(note above that the invocation of the script is surrounded by backticks, to cause the shell to execute the output of the script)
Answering my own question here, using the answers above: if I have more than one related variable to export which use the same value as part of each export, I can do this:
#!/bin/bash
export TEST_EXPORT=$1
export TEST_EXPORT_2=$1_2
export TEST_EXPORT_TWICE=$1_$1
and save as e.g. ~/Desktop/TEST_EXPORTING
and finally $chmod +x ~/Desktop/TEST_EXPORTING
--
After that, running it with source ~/Desktop/TEST_EXPORTING bob
and then checking with export | grep bob should show what you expect.
Exporting a variable into the environment only makes that variable visible to child processes. There is no way for a child to modify the environment of its parent.
Another way you can do it (to steal/expound upon the idea above), is to put the script in ~/bin and make sure ~/bin is in your PATH. Then you can access your variable globally. This is just an example I use to compile my Go source code which needs the GOPATH variable to point to the current directory (assuming you're in the directory you need to compile your source code from):
From ~/bin/GOPATH:
#!/bin/bash
echo declare -x GOPATH=$(pwd)
Then you just do:
#> $(GOPATH)
So you can now use $(GOPATH) from within your other scripts too, such as custom build scripts which can automatically invoke this variable and declare it on the fly thanks to $(pwd).
script1.sh
shell_ppid=$PPID
shell_epoch=$(grep se.exec_start "/proc/${shell_ppid}/sched" | sed 's/[[:space:]]//g' | cut -f2 -d: | cut -f1 -d.)
now_epoch=$(($(date +%s%N)/1000000))
shell_start=$(( (now_epoch - shell_epoch)/1000 ))
env_md5=$(md5sum <<<"${shell_ppid}-${shell_start}"| sed 's/[[:space:]]//g' | cut -f1 -d-)
tmp_dir="/tmp/ToD-env-${env_md5}"
mkdir -p "${tmp_dir}"
ENV_PROPS="${tmp_dir}/.env"
echo "FOO=BAR" > "${ENV_PROPS}"
script2.sh
shell_ppid=$PPID
shell_epoch=$(grep se.exec_start "/proc/${shell_ppid}/sched" | sed 's/[[:space:]]//g' | cut -f2 -d: | cut -f1 -d.)
now_epoch=$(($(date +%s%N)/1000000))
shell_start=$(( (now_epoch - shell_epoch)/1000 ))
env_md5=$(md5sum <<<"${shell_ppid}-${shell_start}"| sed 's/[[:space:]]//g' | cut -f1 -d-)
tmp_dir="/tmp/ToD-env-${env_md5}"
mkdir -p "${tmp_dir}"
ENV_PROPS="${tmp_dir}/.env"
source "${ENV_PROPS}"
echo $FOO
./script1.sh
./script2.sh
BAR
It persists for the scripts run in the same parent shell, and it prevents collisions.
To get this code to run properly, I created a txt file named new_user.txt with the following format (supposed to follow /etc/passwd)
doejjan:x:Doe, Jane Joe+111222:home/STUDENTS/teststu:/bin/bash
smidjoh:x:Smith, John Jay+222333:home/STUDENTS/teststu:/bin/bash
I want to try to display the command that was created to show every record on the screen, below is what I have so far:
#!/bin/bash
while read -r line || [[ -n "$line" ]]; do
username=$(echo "$line" | cut -d: -f1)
GECOS=$(echo "$line" | cut -d: -f5)
homedir=$(echo "$line" | cut -d: -f6)
echo "adduser -g '$GECOS' -d '$homedir' -s /bin/bash '$username'"
done < "$new_user.txt"
I'm getting the error in line 7 that says the following:
.txt:No such file or directory
Can you help me try to fix the error message? Thank you in advance.
From the error message, you can understand that the variable new_user must be empty. Indeed you never assign a value to this variable.
From your description, it follows that $new_user should expand to the value new_user. Say your script is called my_script. If you run it as
new_user=new_user my_script
the error will be gone. If the script is run most of the time on the file new_user.txt, you can - in your script - provide a default value for this variable:
: ${new_user:=new_user}
If you then run it as
my_script
it will pick up new_user.txt, but if you run it by
new_user=old_user my_script
it will run on old_user.txt.
BTW, I personally would prefer passing the file name to the script either via stdin or on the command line, but you have choosen to use a variable for this task, and you can do this of course, if you prefer.
I've placed a bash file inside .zshrc and tried all different ways to run it every time I open a new terminal window or source .zshrc but no luck.
FYI: it was working fine on .bashrc
here is .zshrc script:
#Check if ampps is running
bash ~/ampps_runner.sh & disown
Different approach:
#Check if ampps is running
sh ~/ampps_runner.sh & disown
Another approach:
#Check if ampps is running
% ~/ampps_runner.sh & disown
All the above approaches didn't work (meaning it supposes to run an app named ampps but it doesn't in zsh.
Note: It was working fine before switching to zsh from bash. so it does not have permission or syntax problems.
Update: content of ampps_runner.sh
#! /usr/bin/env
echo "########################"
echo "Checking for ampps server to be running:"
check=$(pgrep -f "/usr/local/ampps" )
#[ -z "$check" ] && echo "Empty: Yes" || echo "Empty: No"
if [ -z "$check" ]; then
echo "It's not running!"
cd /usr/local/ampps
echo password | sudo -S ./Ampps
else
echo "It's running ..."
fi
(1) I believe ~/.ampps_runner.sh is a bash script, so, its first line should be
#!/bin/bash
or
#!/usr/bin/bash
not
#! /usr/bin/env
(2) Then, the call in zsh script (~/.zshrc) should be:
~/ampps_runner.sh
(3) Note: ~/.ampps_runner.sh should be executable. Change it to executable:
$ chmod +x ~/ampps_runner.sh
The easiest way to run bash temporarily from a zsh terminal is to
exec bash
or just
bash
Then you can run commands you previously could only run in bash. An example
help exec
To exit
exit
Now you are back in your original shell
If you want to know your default shell
echo $SHELL
or
set | grep SHELL=
If you want to reliably know your current shell
ps -p $$
Or if you want just the shell name you might use
ps -p $$ | awk "NR==2" | awk '{ print $4 }' | tr -d '-'
And you might just put that last one in a function for later, just know that it is only available if it was sourced in a current shell.
whichShell(){
local defaultShell=$(echo $SHELL | tr -d '/bin/')
echo "Default: $defaultShell"
local currentShell=$(ps -p $$ | awk "NR==2" | awk '{ print $4 }' | tr -d '-')
echo "Current: $currentShell"
}
Call the method to see your results
whichShell
Let's assume I have 3 shell scripts:
script_1.sh
#!/bin/bash
./script_3.sh
script_2.sh
#!/bin/bash
./script_3.sh
the problem is that in script_3.sh I want to know the name of the caller script.
so that I can respond differently to each caller I support
please don't assume I'm asking about $0 cause $0 will echo script_3 every time no matter who is the caller
here is an example input with expected output
./script_1.sh should echo script_1
./script_2.sh should echo script_2
./script_3.sh should echo user_name or root or anything to distinguish between the 3 cases?
Is that possible? and if possible, how can it be done?
this is going to be added to a rm modified script... so when I call rm it do something and when git or any other CLI tool use rm it is not affected by the modification
Based on #user3100381's answer, here's a much simpler command to get the same thing which I believe should be fairly portable:
PARENT_COMMAND=$(ps -o comm= $PPID)
Replace comm= with args= to get the full command line (command + arguments). The = alone is used to suppress the headers.
See: http://pubs.opengroup.org/onlinepubs/009604499/utilities/ps.html
In case you are sourceing instead of calling/executing the script there is no new process forked and thus the solutions with ps won't work reliably.
Use bash built-in caller in that case.
$ cat h.sh
#! /bin/bash
function warn_me() {
echo "$#"
caller
}
$
$ cat g.sh
#!/bin/bash
source h.sh
warn_me "Error: You did not do something"
$
$ . g.sh
Error: You did not do something
g.sh
$
Source
The $PPID variable holds the parent process ID. So you could parse the output from ps to get the command.
#!/bin/bash
PARENT_COMMAND=$(ps $PPID | tail -n 1 | awk "{print \$5}")
Based on #J.L.answer, with more in depth explanations, that works for linux :
cat /proc/$PPID/comm
gives you the name of the command of the parent pid
If you prefer the command with all options, then :
cat /proc/$PPID/cmdline
explanations :
$PPID is defined by the shell, it's the pid of the parent processes
in /proc/, you have some dirs with the pid of each process (linux). Then, if you cat /proc/$PPID/comm, you echo the command name of the PID
Check man proc
Couple of useful files things kept in /proc/$PPID here
/proc/*some_process_id*/exe A symlink to the last executed command under *some_process_id*
/proc/*some_process_id*/cmdline A file containing the last executed command under *some_process_id* and null-byte separated arguments
So a slight simplification.
sed 's/\x0/ /g' "/proc/$PPID/cmdline"
If you have /proc:
$(cat /proc/$PPID/comm)
Declare this:
PARENT_NAME=`ps -ocomm --no-header $PPID`
Thus you'll get a nice variable $PARENT_NAME that holds the parent's name.
You can simply use the command below to avoid calling cut/awk/sed:
ps --no-headers -o command $PPID
If you only want the parent and none of the subsequent processes, you can use:
ps --no-headers -o command $PPID | cut -d' ' -f1
You could pass in a variable to script_3.sh to determine how to respond...
script_1.sh
#!/bin/bash
./script_3.sh script1
script_2.sh
#!/bin/bash
./script_3.sh script2
script_3.sh
#!/bin/bash
if [ $1 == 'script1' ] ; then
echo "we were called from script1!"
elsif [ $1 == 'script2' ] ; then
echo "we were called from script2!"
fi
Below are my two script codes:
**a.sh**
#!/bin/sh
SRC_PATH="/xx/xxxx"
HOST='ftp.xxx.xxx.com'
USER='xxxx'
PASS='xxx'
FTP_SRC_PATH='out'
**b.sh**
#!/bin/sh
/xx/xxx/a.sh
ftp -n $HOST <<END_SCRIPT
quote USER $USER
quote PASS $PASS
binary
prompt off
cd $FTP_SRC_PATH
lcd $SRC_PATH
mget IMS_*.ZIP
bye
END_SCRIPT
My issue is when I am running b.sh , it is not calling a.sh and using the variables defined in it to connect to ftp server.
I have seen many solutions already online but things doesn't seem to work.
Please help
You can include a.shwith the dot command:
. /xx/xxx/a.sh
You can call the script a from within b as follows:
sh ./a.sh
Make sure both the shell scripts are in same file path. Also, you can pass the value within scripts as follows:
sh ./a.sh $d $e
The values passed here can be accessed as $1 $2.
Here is an example:
a.sh:
!/bin/bash
yourvalue="test"
sh ./b.sh $yourvalue
and I'm able to access the variable passed here in the shell script b as $1.