bash command arguments - bash

I have a bash script that, for reasons I won't discuss, cannot be made executable. However, I need to pass arguments to that script.
I have tried this:
bash MyBashScript.sh MyArgumentOne
But the argument MyArgumentOne isn't passed to the script. I know there must be a way to do this, can anyone help?

Your given command should work. Try to debug with calling trough
strace -o all_system_calls.txt -f -ff bash MyBashScript.sh MyArgumentOne
one of the all_system_calls.txt.<pid> files created should contain something like
execve("/bin/bash", ["bash", "MyBashScript.sh", "MyArgumentOne"], [/* 71 vars */]) = 0
If so you know for sure that the argument is passed into your script.

Related

Bindsym does not execute i3wm command

This shortcut does not work in i3wm. It's supposed to show the window list of open apps.
Nothing visible happens, when keyboard shortcut is pressed.
bindsym $mod+space exec bash -c "/home/george/./dmenu-i3-window-jumper.sh"
However the script runs fine from terminal.
The bash code for the script:
https://github.com/minos-org/minos-desktop-tools/blob/master/tools/dmenu-i3-window-jumper
This is a two side issue
First some small config stuff:
I think you got an extra dot in there as ./ in that context just represents the folder preceding it (i.e: /home/george)
You can use the $HOME variable as a stand in for your home folder, i3 will pick it up
I would argue there is really no need for the bash -c, since your file has both a .sh extension and a #!/bin/sh header on the first line, which means you just need to give it execution permissions with chmod +x and it will run with bash anyways.
So in synthesis, you gotta
chmod +x /home/george/dmenu-i3-window-jumper.sh
so the script can be run without calling bash directly,
and your bindsym could be simplified to
bindsym $mod+space exec "$HOME/dmenu-i3-window-jumper.sh"
And then there is the script stuff:
You see, around line 44 the script checks to see if the STDIN is in a terminal, if its not then it tries to pipe a file to the arg array
if [ ! -t 0 ]; then
#add input comming from pipe or file to $#
set -- "${#}" $(cat)
fi
This seems to be the main problem, since you're not running the command in a terminal and you're not giving it a file either.
Your options are A: changing the if so it will always pass an empty string to the argument array
if [ ! -t 0 ]; then
#add input comming from pipe or file to $#
set -- "${#}" ""
fi
or B: create a dummy file with touch ~/dummy and then pass it to the script on the bindsym
bindsym $mod+space exec "$HOME/dmenu-i3-window-jumper.sh < $HOME/dummy"
Both seem to work fine on my setup, good luck!

Problem handling enviroment variable when launching terminal from bash script

The following script gets called with an enviroment variable setted.
I need to launch a terminal and inside that terminal read that variable from another script ( script.sh ).
xfce4-terminal -x sh -c \
"export VAR='${VAR}'
/home/usr/scripts/script.sh"
It works but not when VAR has single quotes in it.
I also feel like there is a better way to pass enviroment variable to the terminal but I don't know how.
I really appreciate any kind of help and I'm sorry for my english.
One of the intended features of the environment is that you can add to it, but you never remove things from it. Add VAR to the current environment, and it will be inherited by xfce4-terminal and any process started by that terminal.
export VAR
xfce4-terminal -x sh -c /home/usr/scripts/script.sh
If you don't want it in the current environment, only in the new terminal's, then use a precommend assignment.
VAR="$VAR" xfce4-terminal -x sh -c /home/usr/scripts/script.sh
This avoids any fragile dynamic script construction like you are contending with.
Since xfce4-terminal appears to not fork a new process itself, I would pass the desired value as an argument to sh.
xfce4-terminal -x sh -c 'VAR="$1" /home/usr/scripts/script.sh' _ "$VAR"
The argument to -c is still a fixed string rather than one generated by interpolating the value of $VAR.

Script not working as Command line

i've created simple bash script that do the following
:
#!/usr/bin/env bash
cf ssh "$1"
When I run the command line from the CLI like cf ssh myapp its running as expected, but when I run the script like
. myscript.sh myapp
I got error: App not found
I dont understand what is the difference, I've provided the app name after I invoke the script , what could be missing here ?
update
when I run the script with the following its working, any idea why the "$1" is not working ...
#!/usr/bin/env bash
cf ssh myapp
When you do this:
. myscript.sh myapp
You don't run the script, but you source the file named in the first argument. Sourcing means reading the file, so it's as if the lines in the file were typed on the command line. In your case what happens is this:
myscript.sh is treates as the file to source and the myapp argument is ignored.
This line is treated as a comment and skipped.
#!/usr/bin/env bash
This line:
cf ssh "$1"
is read as it stands. "$1" takes the value of $1 in the calling shell. Possibly - most likely in your case - it's blank.
Now you should know why it works as expected when you source this version of your script:
#!/usr/bin/env bash
cf ssh myapp
There's no $1 to resolve, so everything goes smoothly.
To run the script and be able to pass arguments to it, you need to make the file executable and then execute it (as opposed to sourcing). You can execute the script for example this way:
./script.bash arg1 arg2

Unable to pass parameters to a perl script inside a bash script

I would like to pass parameters to a perl script using positional parameters inside a bash script "tablecheck.sh". I am using an alias "tablecheck" to call "tablecheck.sh".
#!/bin/bash
/scripts/tables.pl /var/lib/mysql/$1/ /var/mysql/$1/mysql.sock > /tmp/chktables_$1 2>&1 &
Perl script by itself works fine. But when I do "tablecheck MySQLinstance", $1 stays $1. It won't get replaced by the instance. So I get the output as follows:
Exit /scripts/tables.pl /var/lib/mysql/$1/ /var/mysql/$1/mysql.sock > /tmp/chktables_$1 2>&1 &
The job exits.
FYI: alias tablecheck='. pathtobashscript/tablecheck.sh'
I have a bunch of aliases in another bash script. Hence . command.
Could anyone help me... I have gone till the 3rd page of Google to find an answer. Tried so many things with no luck.
I am a noob. But may be it has something to do with it being a background job or $1 in a path... I don't understand why the $1 won't get replaced...
If I copy your exact set up (which I agree with other commenters, is some what unusual) then I believe I am getting the same error message
$ tablecheck foo
[1]+ Exit 127 /scripts/tables.pl /var/lib/mysql/$1/ /var/mysql/$1/mysql.sock > /tmp/chktables_$1 2>&1
In the /tmp/chktables_foo file that it makes there is an additional error message, in my case "bash: /scripts/tables.pl: No such file or directory"
I suspect permissions are wrong in your case

Shell: How to call one shell script from another shell script?

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.

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