[SOLVED] How-to supress output when exit code not 0 ?

Linux_Kidd · 10-21-2011, 12:07 PM

rhel 5.7 in bash

i have a script taht has this, and i need to supress stdout for all $? codes.

passwd -S $1 >/dev/null

works fine when $?=0 but does not bit buck when $? != 0

is this because error message comes from pam?

r.osmanov · 10-21-2011, 12:12 PM

Quote:

Originally Posted by Linux_Kidd

rhel 5.7 in bash
passwd -S $1 >/dev/null

Code:

passwd -S $1 > /dev/null 2>&1

?

crts · 10-21-2011, 12:44 PM

At least in bash you could also do:

Code:

passwd -S $1 &> /dev/null

Which will redirect stdout and stderr to /dev/null.

Linux_Kidd · 10-21-2011, 12:54 PM

Quote:

Originally Posted by crts

At least in bash you could also do:

Code:

passwd -S $1 &> /dev/null

Which will redirect stdout and stderr to /dev/null.

hmmm, yeah, was missing the &

almost there, one last issue. i need to get the stdout or stderr into a variable while at same time supressing screen output

VALID=`passwd -S $1`

i need a variable string to work with while at the same time i need to supress "passwd: unknown user name" , etc. can this be done in a single line?

thnx.

r.osmanov · 10-21-2011, 01:50 PM

Quote:

Originally Posted by Linux_Kidd

hmmm, yeah, was missing the &

almost there, one last issue. i need to get the stdout or stderr into a variable while at same time supressing screen output

VALID=`passwd -S $1`

i need a variable string to work with while at the same time i need to supress "passwd: unknown user name" , etc. can this be done in a single line?

thnx.

Do you want suppress stderr(descriptor no. 2) and get stdout(descriptor no. 1) into a variable?
What's about

Code:

VALID=`passwd -S $1 2>/dev/null`

?

Linux_Kidd · 10-21-2011, 06:17 PM

Quote:

Originally Posted by r.osmanov

Do you want suppress stderr(descriptor no. 2) and get stdout(descriptor no. 1) into a variable?
What's about

Code:

VALID=`passwd -S $1 2>/dev/null`

?

i would any output into the variable and suppress anything to the screen, etc.
thnx

crts · 10-22-2011, 04:26 AM

Quote:

Originally Posted by Linux_Kidd

i would any output into the variable and suppress anything to the screen, etc.
thnx

In this case you can do:

Code:

variable=$(passwd -S users 2>&1)

This redirects stderr to stdout and then assigns the output to variable. However, if you want to check if the command succeeded it is easier to check the exit code with
$?
rather than parsing the returned message in 'variable'.

Linux_Kidd · 10-22-2011, 07:50 AM

Quote:

Originally Posted by crts

In this case you can do:

Code:

variable=$(passwd -S users 2>&1)

This redirects stderr to stdout and then assigns the output to variable. However, if you want to check if the command succeeded it is easier to check the exit code with
$?
rather than parsing the returned message in 'variable'.

i'll give this try. and yes, i will be checking $?, and i also need the string for other parts of script, etc. thanks.

MTK358 · 10-22-2011, 09:32 AM

Also, note that output in stderr and a non-zero exit code are completely unrelated.

Linux_Kidd · 10-24-2011, 07:41 AM

Quote:

Originally Posted by crts

In this case you can do:

Code:

variable=$(passwd -S users 2>&1)

This redirects stderr to stdout and then assigns the output to variable. However, if you want to check if the command succeeded it is easier to check the exit code with
$?
rather than parsing the returned message in 'variable'.

here's the issue, "variable=$(passwd -S $1 2>&1)" always returns $?=0. i was looking for a short one-liner that could fill the variable while at the same time be able to grab $? of the passwd command. i'm no bash expert, but i assume in this case there is a $? for "$(passwd -S $1 2>&1)" and then that gets overwritten when the shell pipes the data into the variable..? doing what i need in two lines is ok, i was just looking for a one-liner, etc.

crts · 10-24-2011, 08:02 AM

Quote:

Originally Posted by Linux_Kidd

here's the issue, "variable=$(passwd -S $1 2>&1)" always returns $?=0. i was looking for a short one-liner that could fill the variable while at the same time be able to grab $? of the passwd command. i'm no bash expert, but i assume in this case there is a $? for "$(passwd -S $1 2>&1)" and then that gets overwritten when the shell pipes the data into the variable..? doing what i need in two lines is ok, i was just looking for a one-liner, etc.

No, it does not - at least not in my bash (GNU bash, version 4.1.5).

Code:

$ passwd -S no-user 2>&1
passwd: user 'no-user' does not exist
$ variable=$(passwd -S no-user 2>&1)
$ echo $?
1

The result is also not piped into the variable. It is assigned. Post the exact code that you are using. Also keep in mind, that $? will hold the exit status of the last executed command. If you have a command after the assignment then $? will hold that commands exit status.

Linux_Kidd · 10-24-2011, 08:29 AM

crts,
indeed you are correct, not sure exactly why i thought my exit codes were always zero (was testing it about 10 times) and thats why i posted "doesnt work for me". but in fact it does work as desired.

thnx, i'll mark this one solved.