CODEWITHSUNDEEP
linuxbash
liv2hak
liv2hak

Reputation: 15010

background processes and outputting to file

I have a bash script that does the following.

spinner()
{
    local pid=$1
    local delay=0.75
    local spinstr='\|/-'
    while [ "$(ps a | awk '{print $1}' | grep $pid)" ]; do
        local temp=${spinstr#?}
        printf " [%c]  " "$spinstr"
        local spinstr=$temp${spinstr%"$temp"}
        sleep $delay
        printf "\b\b\b\b\b\b"
    done
    printf "    \b\b\b\b"
}

printf "Testing for timestamps,please wait..." | tee test_report.txt

(process_1 -f $FILENAME test_1 >> test_report.txt 2>&1)&
spinner $!

printf "\n"

printf "Testing for accurate seq numbers,please wait..." | tee test_report.txt

(process_2 -f $FILENAME sequence >> test_report.txt 2>&1) &
spinner $!
printf "\n";

However when I open the test_report.txt I see only the output of 'process_2' and can't see the output of 'process_1'

I guess the question really is what happens to a the output file that contains the stdout of a background process.?

Upvotes: 0

Views: 299

Answers (1)

William Pursell
William Pursell

Reputation: 212574

The tee test_report.txt is overwriting it. Try tee -a test_report.txt

Upvotes: 4

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