In a bash script I want to get the name of the last command executed in terminal and store it in the variable for later use. I know that !:0 doesn't work in bash script, and I'm looking for some replacement of it.
For example:
#user enters pwd
> pwd
/home/paul
#I call my script and it show the last command
> ./last_command
pwd
this didn't help, it just prints empty line.
getting last executed command from script
Tell the shell to continuously append commands to the history file:
export PROMPT_COMMAND="history -a"
Put the following into your script:
#!/bin/bash
echo "Your command was:"
tail -n 1 ~/.bash_history
as far as I benefit the working one in my .bashrc;
export HISTCONTROL=ignoredups:erasedups
then do this, on console or in script respectively
history 2
cm=$(history 1)
I used to work with UNIX a couple years ago, and I am just starting to get back into it again. I was wondering if anyone could help me with a question.
For example, if I am in bash, I say chsh --shell /bin/tcsh after this I am prompted to enter my password. If I try to say echo $SHELL it will not tell me I have changed shells. It still tells me I am in bash, not C shell. So I have to exit and restart. Once I log back it, then it tells I am in C shell.
Is there a more effective method to change shells? One that does not require me having to log in and out?
Thank you in advance.
chsh(1): change your login shell
Once you change your shell with chsh, it should automatically login to that shell every time you open a terminal.
If you want to use a different shell temporary, just run that shell directly: "tcsh", "zsh", etc..
If you want to use a particular shell for a script use shebang "#!".
Example -- The following on the first line of a shell script will ensure the script is run with sh (and you can do this for any shell available on your system):
#!/bin/sh
Always check your current shell by using :
echo $0
That way you will get the exact process ( your current shell ) you are running. If you print $SHELL it will return to you the default shell that will be open when you login to the server which unless that's what you need its not reliable.
ubuntu$ echo $SHELL
/bin/bash
ubuntu$ echo $0
-bash
ubuntu$ sh
\[\e[31m\]\u\[\e[m\]$ echo $SHELL
/bin/bash
\[\e[31m\]\u\[\e[m\]$ echo $0
sh
\[\e[31m\]\u\[\e[m\]$
Regards!
My default shell is bash. I have set some environment variables in my .bashrc file.
I installed a program which use .cshrc file. It contains the path to several cshell scripts.
When I run the following commands in the shell windows it works perfectly :
exec csh
source .cshrc
exec bash
I have tried to put these commands in bash script, unfortunately it didn't work.
is there another way to write a script in order to get the same result as running commands from a shell windows.
I hope my question is now clear
Many thanks for any help
WARNING : don't put the following script in your .bashrc, it will reload bash and so reload .bashrc again and again (stopable with C-c anyway)
Use preferable this script in your kit/CDS stuff startup script. (cadence presumably)
WARNING 2 : if anything in your file2source fails, the whole 'trick' stops.
Call this script : cshWrapper.csh
#! /bin/csh
# to launch using
# exec cshWrapper.csh file2source.sh
source $1
exec $SHELL -i
and launch it using
exec ./cshWrapper.csh file2source.sh
it will : launch csh, source your file and came back to the same parrent bash shell
Example :
$> ps
PID TTY TIME CMD
7065 pts/0 00:00:02 bash
$>exec ./cshWrapper.csh toggle.csh
file sourced
1
$> echo $$
7065
where in my case i use the file toggle.csh
#! /bin/csh
# source ./toggle.csh
if ! $?TOGGLE then
setenv TOGGLE 0
endif
if ($?TOGGLE) then
echo 'file sourced'
if ($TOGGLE == 0) then
setenv TOGGLE 1
else
setenv TOGGLE 0
endif
endif
echo $TOGGLE
Hope it helps
New proposal, since I faced another problem with exec.
exec kills whatever remains in the script, except if you force a fork by using a pipe after it `exec script |cat'. In such case if you have environment variable in the script, they are not spread back to the script itself, which is not what we want. The only solution I found is to use 3 files (let's call them for the example : main.bash that call first.cshrc and second.sh).
#! /bin/bash
#_main.bash_
exec /bin/csh -c "source /path_to_file/cshrc; exec /bin/bash -i -c /path_to_file/second.sh"
# after exec nothing remains (like Attila the Hun)
# the rest of the script is in 'second.sh'
With that manner, i can launch in a single script call, an old cshrc design kit, and still process some bash command after, and finally launch the main program in bash (let say virtuoso)
I have two shell scripts, a.sh and b.sh.
How can I call b.sh from within the shell script a.sh?
There are a couple of different ways you can do this:
Make the other script executable with chmod a+x /path/to/file(Nathan Lilienthal's comment), add the #!/bin/bash line (called shebang) at the top, and the path where the file is to the $PATH environment variable. Then you can call it as a normal command;
Or call it with the source command (which is an alias for .), like this:
source /path/to/script
Or use the bash command to execute it, like:
/bin/bash /path/to/script
The first and third approaches execute the script as another process, so variables and functions in the other script will not be accessible.
The second approach executes the script in the first script's process, and pulls in variables and functions from the other script (so they are usable from the calling script).
In the second method, if you are using exit in second script, it will exit the first script as well. Which will not happen in first and third methods.
Check this out.
#!/bin/bash
echo "This script is about to run another script."
sh ./script.sh
echo "This script has just run another script."
There are a couple of ways you can do this. Terminal to execute the script:
#!/bin/bash
SCRIPT_PATH="/path/to/script.sh"
# Here you execute your script
"$SCRIPT_PATH"
# or
. "$SCRIPT_PATH"
# or
source "$SCRIPT_PATH"
# or
bash "$SCRIPT_PATH"
# or
eval '"$SCRIPT_PATH"'
# or
OUTPUT=$("$SCRIPT_PATH")
echo $OUTPUT
# or
OUTPUT=`"$SCRIPT_PATH"`
echo $OUTPUT
# or
("$SCRIPT_PATH")
# or
(exec "$SCRIPT_PATH")
All this is correct for the path with spaces!!!
The answer which I was looking for:
( exec "path/to/script" )
As mentioned, exec replaces the shell without creating a new process. However, we can put it in a subshell, which is done using the parantheses.
EDIT:
Actually ( "path/to/script" ) is enough.
If you have another file in same directory, you can either do:
bash another_script.sh
or
source another_script.sh
or
. another_script.sh
When you use bash instead of source, the script cannot alter environment of the parent script. The . command is POSIX standard while source command is a more readable bash synonym for . (I prefer source over .). If your script resides elsewhere just provide path to that script. Both relative as well as full path should work.
Depends on.
Briefly...
If you want load variables on current console and execute you may use source myshellfile.sh on your code. Example:
#!/bin/bash
set -x
echo "This is an example of run another INTO this session."
source my_lib_of_variables_and_functions.sh
echo "The function internal_function() is defined into my lib."
returned_value=internal_function()
echo $this_is_an_internal_variable
set +x
If you just want to execute a file and the only thing intersting for you is the result, you can do:
#!/bin/bash
set -x
./executing_only.sh
bash i_can_execute_this_way_too.sh
bash or_this_way.sh
set +x
You can use /bin/sh to call or execute another script (via your actual script):
# cat showdate.sh
#!/bin/bash
echo "Date is: `date`"
# cat mainscript.sh
#!/bin/bash
echo "You are login as: `whoami`"
echo "`/bin/sh ./showdate.sh`" # exact path for the script file
The output would be:
# ./mainscript.sh
You are login as: root
Date is: Thu Oct 17 02:56:36 EDT 2013
First you have to include the file you call:
#!/bin/bash
. includes/included_file.sh
then you call your function like this:
#!/bin/bash
my_called_function
Simple source will help you.
For Ex.
#!/bin/bash
echo "My shell_1"
source my_script1.sh
echo "Back in shell_1"
Just add in a line whatever you would have typed in a terminal to execute the script!
e.g.:
#!bin/bash
./myscript.sh &
if the script to be executed is not in same directory, just use the complete path of the script.
e.g.:`/home/user/script-directory/./myscript.sh &
This was what worked for me, this is the content of the main sh script that executes the other one.
#!/bin/bash
source /path/to/other.sh
The top answer suggests adding #!/bin/bash line to the first line of the sub-script being called. But even if you add the shebang, it is much faster* to run a script in a sub-shell and capture the output:
$(source SCRIPT_NAME)
This works when you want to keep running the same interpreter (e.g. from bash to another bash script) and ensures that the shebang line of the sub-script is not executed.
For example:
#!/bin/bash
SUB_SCRIPT=$(mktemp)
echo "#!/bin/bash" > $SUB_SCRIPT
echo 'echo $1' >> $SUB_SCRIPT
chmod +x $SUB_SCRIPT
if [[ $1 == "--source" ]]; then
for X in $(seq 100); do
MODE=$(source $SUB_SCRIPT "source on")
done
else
for X in $(seq 100); do
MODE=$($SUB_SCRIPT "source off")
done
fi
echo $MODE
rm $SUB_SCRIPT
Output:
~ ❯❯❯ time ./test.sh
source off
./test.sh 0.15s user 0.16s system 87% cpu 0.360 total
~ ❯❯❯ time ./test.sh --source
source on
./test.sh --source 0.05s user 0.06s system 95% cpu 0.114 total
* For example when virus or security tools are running on a device it might take an extra 100ms to exec a new process.
pathToShell="/home/praveen/"
chmod a+x $pathToShell"myShell.sh"
sh $pathToShell"myShell.sh"
#!/bin/bash
# Here you define the absolute path of your script
scriptPath="/home/user/pathScript/"
# Name of your script
scriptName="myscript.sh"
# Here you execute your script
$scriptPath/$scriptName
# Result of script execution
result=$?
chmod a+x /path/to/file-to-be-executed
That was the only thing I needed. Once the script to be executed is made executable like this, you (at least in my case) don't need any other extra operation like sh or ./ while you are calling the script.
Thanks to the comment of #Nathan Lilienthal
Assume the new file is "/home/satya/app/app_specific_env" and the file contents are as follows
#!bin/bash
export FAV_NUMBER="2211"
Append this file reference to ~/.bashrc file
source /home/satya/app/app_specific_env
When ever you restart the machine or relogin, try echo $FAV_NUMBER in the terminal. It will output the value.
Just in case if you want to see the effect right away, source ~/.bashrc in the command line.
There are some problems to import functions from other file.
First: You needn't to do this file executable. Better not to do so!
just add
. file
to import all functions. And all of them will be as if they are defined in your file.
Second: You may be define the function with the same name. It will be overwritten. It's bad. You may declare like that
declare -f new_function_name=old_function_name
and only after that do import.
So you may call old function by new name.
Third: You may import only full list of functions defined in file.
If some not needed you may unset them. But if you rewrite your functions after unset they will be lost. But if you set reference to it as described above you may restore after unset with the same name.
Finally In common procedure of import is dangerous and not so simple. Be careful! You may write script to do this more easier and safe.
If you use only part of functions(not all) better split them in different files. Unfortunately this technique not made well in bash. In python for example and some other script languages it's easy and safe. Possible to make partial import only needed functions with its own names. We all want that in next bush versions will be done the same functionality. But now We must write many additional cod so as to do what you want.
Use backticks.
$ ./script-that-consumes-argument.sh `sh script-that-produces-argument.sh`
Then fetch the output of the producer script as an argument on the consumer script.
I'm using this bash script:
for a in `sort -u $HADOOP_HOME/conf/slaves`; do
rsync -e ssh -a "${HADOOP_HOME}/conf" ${a}:"${HADOOP_HOME}"
done
for a in `sort -u $HBASE_HOME/conf/regionservers`; do
rsync -e ssh -a "${HBASE_HOME}/conf" ${a}:"${HBASE_HOME}"
done
When I call this script directly from shell, there are no problems and it works fine. But when I call this script from another script, although the script does its job, I get this message at the end:
sort: open failed: /conf/slaves: No such file or directory
sort: open failed: /conf/regionservers: No such file or directory
I have set $HADOOP_HOME and $HBASE_HOME in /etc/profile and the script does the job right. But I don't understand why it gives this message in the end.
Are you sure it's doing it right? When you call this script from the shell it is acting as an interactive shell which reads and sources /etc/profile and ~/.bash_profile if it exists. When you call it from another script it is running as non-interactive and wont source those files. If you want a non-interactive shell to source a file you can do this by setting the BASH_ENV environment variable.
#!/bin/bash
export BASH_ENV=/etc/profile
./call/to/your/HADOOP/script.sh
Everything points to those variables not being defined when your script runs.
You should ensure that they are set for your script. Before the first loop, place the line:
echo "[${HADOOP_HOME}] [${HBASE_HOME}]"
and make sure that doesn't output "[] []" (or even one "[]").
Additionally, put a set +x line at the top of the script - this will output lines before executing them and you can see what's being done.
Keep in mind that some shells don't pass on environment variables to subshells unless you explicitly export them (setting them is not enough).