Bash: returning line numbers from a function

Question

I am trying to implement an abort function in a bash script, currently it looks like this:

function abort {
  LOGFILE=/var/log/bash_test
  DATE=$(date "+%Y %b %a %H:%M:%S")
  printf "\n------| ERROR: %s line %03d in %s |------\n" "$DATE" "$2" "${0}" >> "$LOGFILE"
  echo "$1" >> "$LOGFILE"
}

abort "Some kind of error..." $LINENO

This will produce this in /var/log/bash_test

------| ERROR: 2014 Jan Tue 12:50:12 line 007 in /home/user/test.sh |------
Some kind of error

My question is this: The method of giving the line number manually (eg. $2 < $LINENO) is ugly and repetitive. Is there a way to make this function detect the $LINENO from outside the function automatically? So that i can give the command

abort "Some kind of error..."

And still get the line number?

Answer 1

You should be able to use BASH_LINENO:

$ cat test.sh 
my_environment() {
    echo "Stack size: ${#BASH_LINENO[@]}"
    echo "Caller line: ${BASH_LINENO[$((${#BASH_LINENO[@]} - 2))]}"
}
my_environment
$ sh test.sh 
Stack size: 2
Caller line: 5

That is, the second to last entry will be the line number where the call to my_environment happened.

From the man bash section about BASH_LINENO:

An array variable whose members are the line numbers in source files where each corresponding member of FUNCNAME was invoked. ${BASH_LINENO[$i]} is the line number in the source file (${BASH_SOURCE[$i+1]}) where ${FUNCNAME[$i]} was called (or ${BASH_LINENO[$i-1]} if referenced within another shell function). Use LINENO to obtain the current line number.

Answer 2

with awk, you can get the line number. Below command will search the key word "ERROR" first, and export the line number.

awk '/ERROR/{print NR}' logfile