I am wondering if there is a way to write set -x output to a file from within a script, rather than call a tee from the command prompt.
For example, I usually use myScript.sh 2>&1 | tee mylog.log form the command prompt. This copies the set -x as I expect to the log file.
Is there a way to internalise this within myScript.sh so I can set it as a flag to be turned off if I do not need to debug. running only myScript.sh from the command prompt.
thx
Art
Inside your script place this line at top:
#!/bin/bash
[[ $1 == "debug" ]] && { exec 2>err.txt; set -x; }
# rest of the script
Now when you call your script as:
./myScript.sh debug
You will get a file created as err.txt containing output of set -x (with other error, if any)
Related
In reference to https://stackoverflow.com/a/11886837/1996022 (also shamelessly stole the title) where the question is how to capture the script's output I would like to know how I can additionally capture the scripts input. Mainly so scripts that also have user input produce complete logs.
I tried things like
exec 3< <(tee -ia foo.log <&3)
exec <&3 <(tee -ia foo.log <&3)
But nothing seems to work. I'm probably just missing something.
Maybe it'd be easier to use the script command? You could either have your users run the script with script directly, or do something kind of funky like this:
#!/bin/bash
main() {
read -r -p "Input string: "
echo "User input: $REPLY"
}
if [ "$1" = "--log" ]; then
# If the first argument is "--log", shift the arg
# out and run main
shift
main "$#"
else
# If run without log, re-run this script within a
# script command so all script I/O is logged
script -q -c "$0 --log $*" test.log
fi
Unfortunately, you can't pass a function to script -c which is why the double-call is necessary in this method.
If it's acceptable to have two scripts, you could also have a user-facing script that just calls the non-user-facing script with script:
script_for_users.sh
--------------------
#!/bin/sh
script -q -c "/path/to/real_script.sh" <log path>
real_script.sh
---------------
#!/bin/sh
<Normal business logic>
It's simpler:
#! /bin/bash
tee ~/log | your_script
The wonderful thing is your_script can be a function, command or a {} command block!
I have an unix shell script. I have put -x in shell to see all the execution step. Now I want to capture these in one log file on a daily basis.
Psb script.
#!/bin/ksh -x
Logfile= path.log.date
Print " copying file" | tee $logifle
Scp -i key source destination | tee -a $logfile.
Exit 0;
First line of the shell script is known as shebang , which indicates what interpreter has to be execute the below script.
Similarly first line is commented which denotes coming lines not related to that interpreted session.
To capture the output, run the script redirect your output while running the script.
ksh -x scriptname >> output_file
Note:it will output what your script's doing line by line
There are two cases, using ksh as your shell, then you need to do IO redirection accordingly, and using some other shell and executing a .ksh script, then IO redirection could be done based on that shell. Following method should work for most of the shells.
$ cat somescript.ksh
#!/bin/ksh -x
printf "Copy file \n";
printf "Do something else \n";
Run it:
$ ./somescript.ksh 1>some.log 2>&1
some.log will contain,
+ printf 'Copy file \n'
Copy file
+ printf 'Do something else \n'
Do something else
In your case, no need to specify logfile and/or tee. Script would look something like this,
#!/bin/ksh -x
printf "copying file\n"
scp -i key user#server /path/to/file
exit 0
Run it:
$ ./myscript 1>/path/to/logfile 2>&1
2>&1 captures both stderr and stdout into stdout and 1>logfile prints it out into logfile.
I would prefer to explicitly redirecting the output (including stderr 2> because set -x sends output to stderr).
This keeps the shebang short and you don't have to cram the redirecton and filename-building into it.
#!/bin/ksh
logfile=path.log.date
exec >> $logfile 2>&1 # redirecting all output to logfile (appending)
set -x # switch on debugging
# now start working
echo "print something"
Our shell script contains the header
#!/bin/bash -x
that causes the commands to also be listed. Instead of having to type
$ ./script.sh &> log.txt
I would like to add a command to this script that will log all following output (also) to a log file. How this is possible?
You can place this line at the start of your script:
# redirect stdout/stderr to a file
exec &> log.txt
EDIT: As per comments below:
#!/bin/bash -x
# redirect stdout/stderr to a file and still show them on terminal
exec &> >(tee log.txt; exit)
I scheduled a script using at scheduler in linux.
The job ran fine but the echo statements which I had redirected to a file are no where to be found.
The at scheduling command is as follows:
at -f /app/data/scripts/func_test.sh >> /app/data/log/log.txt 2>&1 -v 09:50
Can anyone point out what is the issue with the above command.
I cannot see any echo statements from the script in the log.txt file
To include shell syntax like I/O redirection, you'll need to either fold it into your script, or pass the input to at via standard input, like so:
at -v 09:50 <<EOF
sh /app/data/scripts/func_test.sh >> /app/data/log/log.txt 2>&1
EOF
If func_test.sh is already executable, you can omit the sh from the beginning of the command; it's there to ensure that you are passing a valid command line to at.
You can also simply ensure that your script itself redirects all its output to a specific log file. As an example,
#!/bin/bash
echo foo
echo bar
becomes
#!/bin/bash
{
echo foo
echo bar
} >> /app/data/log/log.txt 2>&1
Then you can simply run your script with at using
at -f /app/data/scripts/func_test.sh -v 09:50
with no output redirection, because the script itself already redirects all its output to that file.
I need to check the run of a bash script, with a source call, something like:
#!/bin/bash
some code here
source script_b.sh
more code here
I run:
$bash -x script_a.sh
and I get,
+ some echo here
+ script_b.sh
+ some more echo here
But all echoes are from script_a.sh. All the code from script_b.sh is hidden, so I can not trace what is really happening.
Is there any way I can check the execution of script_b.sh within script_a.sh?
You could try "bash -x script_b.sh" inside of the parent script.
Edit:
This worked for me. If you run the parent script with bash -x you will see everything for both. "set -x" will set the debug flag for the environment...in the script? I'm not sure, and fifo is still magic to me.
echo "start of script"
set -x
mkfifo fifo
cat /dev/null < fifo | fifo > source .bash_profile
rm fifo
echo "end of script"