As an example, I am trying to capture the raw commands that are output by the following script:
https://github.com/adampointer/go-deribit/blob/master/scripts/generate-models.sh
I have tried to following a previous answer:
BASH: echoing the last command run
but the output I am getting is as follows:
last command is gojson -forcefloats -name="${struct}" -tags=json,mapstructure -pkg=${p} >> models/${p}/${name%.*}_request.go
What I would like to do is capture the raw command, in other words have variables such as ${struct}, ${p} and ${p}/${name%.*} replaced by the actual values that were used.
How do I do this?
At the top of the script after the hashbang #!/usr/bin/env bash or #!/bin/bash (if there is any) add set -x
set -x Print commands and their arguments as they are executed
Run the script in debug mode which will trace all the commands in the script: https://stackoverflow.com/a/10107170/988525.
You can do that without editing the script by typing "bash generate-models.sh -x".
I have a bash script start.sh which calls another run.sh, which takes me to another prompt where I have to delete a file file.txt and then exit out of that prompt.
When I call run.sh from inside start.sh, I see the prompt and I believe that it deletes the file.txt but the inner/new prompt waits for me to exit out of it while the script is running - meaning it needs intervention to proceed. How do I avoid it in bash?
In Python I can use Popen and get it going but not sure about bash.
EDIT: I would rather like to know what command to provide to exit out of the shell (generated from running run.sh") so I can go back to the prompt where "start.sh" was started.
Etan: To answer your question
VirtualBox:~/Desktop/ > ./start
company#4d6z74d:~$ ->this is the new shell
company#4d6z74d:~$ logout ---> I did a "Control D here" so the script could continue.
Relevant part of start.sh which:
/../../../../run.sh (this is the one that takes us to the new $ prompt)
echo "Delete file.txt "
rm -f abc/def/file.txt
You can run run.sh in the background using &. In start.sh, you would invoke the script via /path/run.sh &. Now, start.sh will exit without waiting for run.sh to finish (which is running in the background).
There is a RESLOC variable in my .profile file, which changes from time to time. So i wrote a script just take input from user on the new name.
cat tst.sh
echo "Enter the Result Location name where you would like your results to go."
read RESL
perl -pi.bak -e "s/([\s]+)RESLOC=\/result\/([\S]+)/$1 RESLOC=\/result\/${RESL}/g" /user/.profile
cd /user
. /user/.profile
echo "$RESLOC"
The last echo statement gives the output as the value given by user.
But when i do echo $RESLOC after the script has been executed in the terminal, it displays the old value.
O/P of the script:
Enter the Result Location name where you would like your results to go.
Release12
/user/Release12
When try to display the RESLOC after the execution is complete.
echo $RESLOC
/user/Release11
The .profile file has been updated with Release12. But it is not sourced properly.
Please help.
When you run tst.sh a new shell process is spawned and when it ends your environment will return to the previous instance of the shell, i.e. the one from which you ran tst.sh.
To modify the environment in your current shell, you'll need to source tst.sh;
. tst.sh
This will run tst.sh in your current shell and not spawn a new shell process.
I am running a bash script that takes hours. I was wondering if there is way to monitor what is it doing? like what part of the script is currently running, how long did it take to run the whole script, if it crashes at what line of the script stopped working, etc. I just want to receive feedback from the script. Thanks!!!
from man page for bash,
set -x
After expanding each simple command, for command, case command, select command, or arithmetic for command, display the expanded value of PS4, followed by the command and its expanded arguments or associated word list.
add these to the start of your script,
export PS4='+{${BASH_SOURCE}:$LINENO} '
set -x
Example,
#!/bin/bash
export PS4='+{${BASH_SOURCE}:$LINENO} '
set -x
echo Hello World
Result,
+{helloworld.sh:6} echo Hello World
Hello World
Make a status or log file. For example add this inside your script:
echo $(date) - Ok >> script.log
Or for a real monitoring you can use strace on linux for see system call, example:
$ while true ; do sleep 5 ; done &
[1] 27190
$ strace -p 27190
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.