I'm trying to write a simple script using bash which must take two variables: a name and a command; then apply these two variables inside a "alias" command like so: alias $1=$2.
It`s for an online school activity which teaches introduction to bash scripting.
What I have come up with is:
function doalias {
alias $1=$2
}
doalias $1 $2
Next thing I do is write inside the terminal ./doalias.sh inst "sudo apt-get install".
Now the message I get from the bash terminal is that "=" is not recognized as a command. Why is this and how can I solve it?
You need to put the passed parameter in quotes when actioning the alias command and so:
function doalias {
alias "$1=$2"
}
doalias "$1" "$2"
To have the alias set in the current shell, you will need to source the script and so:
source doalias.sh inst "sudo apt-get install"
I'm trying to get the current shell command (not $history[1] !), specifically i'd like to be able to replace the fish_title - (pwd), with specific running programs i.e. python if I'm in interactive shell, vim etc.
From the documentation:
The first argument to fish_title will contain the most recently executed foreground command as a string, starting with fish 2.2.
So a simple
function fish_title
echo $argv (set -q SSH_CONNECTION; and hostname)":" $PWD
end
should work.
From the documentation, Special Variables
_, the name of the currently running command.
The default fish_title function does this already, does it not?
function fish_title
echo $_ " "
set -q SSH_CONNECTION; and echo (hostname)":"
pwd
end
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 have a script which allows to execute Bash processes in the background, i called it "backy". Programs I want to run in background I call like this:
backy long-running-script param1 param2
The problem is now that I loose the Bash completion for long-running-script if I prepend another script.
I want to write a Bash completion file which preserves not only the Bash completion for long-running-script and all of its parameters, also for every other script that I want to call with backy.
I have some experience with Bash completion, but I'm just missing the command which I can insert into my Bash completion script so that it completes with the completion of the script that is to be called. Any ideas?
My completion so far:
have backy &&
_backy_complete()
{
local cur prev goals
COMPREPLY=()
cur=${COMP_WORDS[COMP_CWORD]}
# How to get the completion from the script that is the param for backy,
# in a generic way?
COMPREPLY=( ????? )
return 0
} &&
complete -F _backy_complete backy
EDIT - SOLUTION:
Thanks to Lekensteyn, I replaced the content of my existing bash completion script with just this line:
complete -F _command backy
There is already a bash_completion function for such cases:
complete -F _command backy
It's used for autocompleting the commands after sudo, fakeroot and others. Any arguments passed to backy are ignored like:
backy --whatever --this --is=ignored not ignored anymore
In my bash script, I execute some commands as another user. I want to call a bash function using su.
my_function()
{
do_something
}
su username -c "my_function"
The above script doesn't work. Of course, my_function is not defined inside su. One idea I have is to put the function into a separate file. Do you have a better idea that avoids making another file?
You can export the function to make it available to the subshell:
export -f my_function
su username -c "my_function"
You could enable 'sudo' in your system, and use that instead.
You must have the function in the same scope where you use it. So either place the function inside the quotes, or put the function to a separate script, which you then run with su -c.
Another way could be making cases and passing a parameter to the executed script.
Example could be:
First make a file called "script.sh".
Then insert this code in it:
#!/bin/sh
my_function() {
echo "this is my function."
}
my_second_function() {
echo "this is my second function."
}
case "$1" in
'do_my_function')
my_function
;;
'do_my_second_function')
my_second_function
;;
*) #default execute
my_function
esac
After adding the above code run these commands to see it in action:
root#shell:/# chmod +x script.sh #This will make the file executable
root#shell:/# ./script.sh #This will run the script without any parameters, triggering the default action.
this is my function.
root#shell:/# ./script.sh do_my_second_function #Executing the script with parameter
this function is my second one.
root#shell:/#
To make this work as you required you'll just need to run
su username -c '/path/to/script.sh do_my_second_function'
and everything should be working fine.
Hope this helps :)