CODEWITHSUNDEEP
bashshell
vidya
vidya

Reputation: 103

Output redirection in shell script not working completely

>cat test.sh
#!bin/bash
testf="/home/vidya/test/"
cd $testf
`make clean all` >/dev/null 2>&1
if [ -e test ]
then
`./test << end
source(scripts/all.src);
quit;
end` >/dev/null 2>&1
echo -e "\nLog Result is:`grep Tests test.log | tail -1`\n"
fi

When I run the above script below is the output. The problem here is even though I used output redirection while executing test case but still it is printing on console like "Opened file .... " for each file. It it redirecting most of the test case execution but not completely. Is there anyway to suppress this output on console ?

[vidya ]$ sh test.sh
Opened file </home/vidya/test/scripts/all.src>
Opened file </home/vidya/test/scripts/file1>
Opened file </home/vidya/test/scripts/file2>
.
.
.
Opened file </home/vidya/test/scripts/file10>

Log Result is:    Tests: 1005 PASSED / 48 FAILED

Upvotes: 3

Views: 2483

Answers (2)

sjsam
sjsam

Reputation: 21965

The legacy back-ticks or modern $() in bash stands for command substitution ie the output of the what is enclosed in the back-ticks will be executed as it is a command

Change

`make clean all` 1>/dev/null 2>&1

to

make clean all >/dev/null 2>&1 # Removed the backticks

Instead of output redirection, you could also think about putting

exec 1>&3 #making a copy file descriptor 1(stdout) to 3.
exec 1>/dev/null 2>&1

in the beginning of test.sh just after the shebang to run the script silently. Then at the point where you need to start printing stuff again, revert the settings

exec 3>&1 Restoring file descriptor 1 to stdout.

Upvotes: 1

user5081924
user5081924

Reputation:

as sjsam has answered remove the backticks . I think the syntax is like this

./test >/dev/null 2>&1 << end
source(scripts/all.src);
quit;
end

Upvotes: 0

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