LinuxQuestions.org - [SOLVED] Print stdout and stderr to screen + log file and catch any errors in shell script

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Print stdout and stderr to screen + log file and catch any errors in shell script

Hello!
I'm writing a shell script that executes a command. I want to save both stdout and stderr to the same log file + print to sceen and at the same time catch if there is an error. What I have so far is:

Code:

set -o pipefail

errorsFound=0

command arg1 arg2 2>&1 | tee logFile.log || errorsFound=1

if [[ $errorsFound == 1 ]]; then 

        echo "command failed!"

        exit 1 

fi

When I run the shell script the logFile.log is created and seems to contain all the output information. However, the variable "errorsFound" always turns to 1, even though there are no errors from the command. Have I set this up wrong?
Thanks in advance!

yes, tee will eat up the error code.
try the following:

Code:

set -o pipefail

errorsFound=0

{

command args 2>&1 || {

    echo command failed

    exit 1

}

} | tee logfile

Incidentally, '==' is a string comparator; you'd want '-eq'
http://tldp.org/LDP/abs/html/comparison-ops.html

Thank you pan64, your solution worked just fine! I used the code below

Code:

set -o pipefail

errorsFound=0

{

command args 2>&1 || { errorsFound=1; }



if [[ $errorsFound == 1 ]]; then { echo "command failed"; exit 1; } fi

} | tee logfile