bash, return value and $? from a command call

a4z · 05-26-2014, 07:19 AM

hi,

when I execute a program in a script I get $? with 0 if it works and 1 if it fails

in a script ,if I do

myvar=$(some_command)
$? is always 0,even if some_command fails,

also with
myvar=`some_command`
$? does not tell me if some_command was successfully
?$ -eq 0.
do I do something wrong?

how can I have the return value of a program and the information if it was successfully

pan64 · 05-26-2014, 07:47 AM

probably this discussion helps a bit: http://www.unixli.com/q/answers-how-...ail-23026.html

rknichols · 05-26-2014, 09:01 AM

Works fine for me

Code:

$ X=$(grep "if" /etc/profile)
$ echo $?
0
$ X=$(grep "plugh" /etc/profile)
$ echo $?
1

That's in bash 4.x. What shell are you using?

NevemTeve · 05-26-2014, 09:27 AM

Works for me, too. Example: set 'erase' if 'kbs' is defined

Code:

X=$(tput kbs) && stty erase "$X"

Habitual · 05-26-2014, 09:50 AM

What is "some_command"?

a4z · 05-27-2014, 02:10 AM

the problem seems to be that I am in a function and use a local variable

Code:

func(){
  local foo=$(sqlite3 doesnotexist "absolut no sql;")
  echo $?
}

func

this prints 0, but I expected 1

NevemTeve · 05-27-2014, 02:56 AM

try in two steps:

Code:

func() {
  local foo
  foo=$(sqlite3 doesnotexist "absolut no sql;")
  echo $?
}

a4z · 05-27-2014, 03:23 AM

Quote:

Originally Posted by NevemTeve

try in two steps:

Code:

func() {
  local foo
  foo=$(sqlite3 doesnotexist "absolut no sql;")
  echo $?
}

this works, basically
I had local -r foo=$(some_command), so I need to remove the -r or use 2 steps
checked also that using declare does change $?

NevemTeve · 05-27-2014, 03:45 AM

if you are dead-set on using 'readonly', then it will be longer:

Code:

func() {
  local tmp
  tmp=$(sqlite3 doesnotexist "absolut no sql;") || { echo $?; exit; }

  local -r foo="$tmp"
  echo 0
}

rtmistler · 05-29-2014, 01:19 PM

Partially doing a double-take here.

(1) Yes it should work as others have said

(2) Why the need to check $? when you've assigned the result to a variable? If I do that, I check "the variable". (That's where I'm doing the double-take.)

(3) When debugging a script, start by adding "set -xv" at the top of the script to see debug information.

Some tips Blog Entry: Bash Scripting for Dummies and Geniuses

suicidaleggroll · 05-29-2014, 01:29 PM

Quote:

Originally Posted by rtmistler

(2) Why the need to check $? when you've assigned the result to a variable? If I do that, I check "the variable". (That's where I'm doing the double-take.)

I'm not sure I get your meaning. The variable will contain the program's stdout. $? will contain the program's exit status. The two are not the same and can not be used interchangeably. Trying to parse a program's stdout to see if an error occurred would be slow, cumbersome, and could even be impossible with certain commands.

rtmistler · 05-29-2014, 01:59 PM

Quote:

Originally Posted by suicidaleggroll

I'm not sure I get your meaning. The variable will contain the program's stdout. $? will contain the program's exit status. The two are not the same and can not be used interchangeably. Trying to parse a program's stdout to see if an error occurred would be slow, cumbersome, and could even be impossible with certain commands.

Sorry, you are correct. I was incorrectly expecting exit status to be stored in that variable. But of course the easiest thing to verify my ignorance was to write a short script and include "set -xv".