CODEWITHSUNDEEP
linuxbashshelllogging
Galet
Galet

Reputation: 6309

How to use different filename to log stdout and stderr in different shell script files

Here are my script files:-

main.sh

# Log stdout and stderr
log_stdout_and_stderr() {
  # Close STDOUT file descriptor
  exec 1<&-
  # Close STDERR FD
  exec 2<&-

  # Open STDOUT as $1 file for read and write.
  exec 1<>$1

  # Redirect STDERR to STDOUT
  exec 2>&1
}

Log a single line

log() {
  echo "[$(date)]: $*"
}

log_stdout_and_stderr main.log

log "Started main.sh"
log "Completed main.sh"

# call first_script
source first_script.sh

first_script.sh

log_stdout_and_stderr first_script.log

log "Started first_script.sh"

# call second_script
source second_script.sh

log "Completed first_script.sh"

second_script.sh

log_stdout_and_stderr second_script.log

log "Started second_script.sh"
log "Completed second_script.sh"

Below are my outputs logs:-

main.log

Started main.sh
Completed main.sh

first_script.log

Started first_script.sh

second_script.log

Started second_script.sh
Completed second_script.sh
Completed first_script.sh

I want to log messages should be logged like below in the log files.

Expected output:-

main.log

Started main.sh
Completed main.sh

first_script.log

Started first_script.sh
Completed first_script.sh

second_script.log

Started second_script.sh
Completed second_script.sh

I am calling second_script.sh file from first_script.sh. I want to store the log messages used by second_script.sh in second_script.log

How can I do it? Let me know if anyone is not clear about the question.

Upvotes: 2

Views: 213

Answers (1)

Rdit
Rdit

Reputation: 1562

You are using source in order to call the scripts from the main script. Sourcing a script means that it is parsed and executed by the current shell itself. It's as if you typed the contents of the script within the main script. i.e.:

The following is an equivalent when you call the second_script.sh within the first_script.sh:

first_script.sh

log_stdout_and_stderr first_script.log
log "Started first_script.sh"

# call second_script
####### second_script.sh called with sourcing #######
log_stdout_and_stderr second_script.log
log "Started second_script.sh"
log "Completed second_script.sh"
######################################################

log "Completed first_script.sh"

Since the script will be executed sequentially, logs will be stored according to the last called log_stdout_and_stderr.

In order to avoid this:

  1. Modify first_script.sh to:
log_stdout_and_stderr first_script.log
log "Started first_script.sh"

#call second_script
source second_script.sh

log_stdout_and_stderr first_script.log
log "Completed first_script.sh"
  1. Or call the another script from within the script using the full path to script:
log_stdout_and_stderr first_script.log

log "Started first_script.sh"

# call second_script 
/path/to/second_script.sh

log "Completed first_script.sh"

This way you'll call another bash shell to execute the other script (in this case second_script.sh). Calling the script this way you'll need to add execution permissions to the script you're calling.

Upvotes: 1

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