Bash function from another script error handler

Question

I have the following situation. Script A source script B. Script B has a function named check, which does no error handling on any of its executed commands.

check () {
  cmd1
  cmd2
  cmd3
  ...
}

My goal is to call the function check from A, exiting if any of cmd* fails.

I have tried from A "check || exit 1" but if cmd2 fails and cmd3 succeeds, A does not exit. set -e did not do the job for me as well.

Any help is appreciated.

set -e seems to work for me. Please, give us more details. — choroba
– choroba, Commented Mar 5, 2016 at 20:44
"If a compound command or shell function executes in a context where -e is being ignored, none of the commands executed within the compound command or function body will be affected by the -e setting, even if -e is set and a command returns a failure status." gnu.org/software/bash/manual/html_node/… — miken32
– miken32, Commented Mar 5, 2016 at 22:39
you can do cmd1 && cmd2 && cmd3 ... in check, "check || exit 1" should work then. — divyum
– divyum, Commented Mar 5, 2016 at 22:56
@miken32 : can you point us to where the "context where -e is being ignored" is described/defined? Thanks and Good luck to all. — shellter
– shellter, Commented Mar 6, 2016 at 4:51

SLePort · Accepted Answer · 2016-03-06 06:32:04Z

2

You can try a custom exit function.

With fileA :

#!/bin/bash
source fileB
check

and fileB :

#!/bin/bash
declare -a exitcode

check() {
  cmd1
  cmd2
  trap finish EXIT
}

cmd1() { 
  myexit 1
}

cmd2() {
  myexit 0
}

myexit() {
  exitcode+=$1
}

finish() {
  [[ ${exitcode[*]} =~ 1 ]] && exit 1  || exit 0
}

Then :

$ ./fileA; echo $?
1

The exit codes are appended in an array that is checked on exit in the finish() function.

answered Mar 6, 2016 at 6:32

SLePort

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Comments

Ali Abdallah · Accepted Answer · 2016-03-06 07:05:30Z

2

I figure it out. The problem was that if I "set -e" and call the function "func || exit 1" then func subcommands will continue even if cmd1 fails. Calling simply "func" does the trick.

answered Mar 6, 2016 at 7:05

Ali Abdallah

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