CODEWITHSUNDEEP
linuxbashpipestdoutstdin
edi9999
edi9999

Reputation: 20554

Output to stdout only when stdin has finished

What is the simplest command to wait for stdin to be finished to output something to stdout.

For example, If I write : 

{
echo "foo"
sleep 1
echo "bar"
} | command

It should show 

foo
bar

after 1 second (and not the first foo at the beginning).

My solution so far is to do twice tac, For example : 

{
echo "foo"
sleep 1
echo "bar"
} | tac | tac

It has the advantage of handling all ANSI escapes/colors well.

Is they any simpler way to do what I want ?

Upvotes: 0

Views: 179

Answers (4)

user2350426
user2350426

Reputation:

Capture output, then echo/print it:

output=$(
    echo "foo"
    sleep 1
    echo "bar"
) 
printf "%s\n" "$output"

Or just:

printf "%s\n" "$(
    echo "foo"
    sleep 1
    echo "bar"
)"

Upvotes: 0

bufh
bufh

Reputation: 3410

Alternatives using bashims:

cat <<EOF
$(
  echo foo
  sleep 3
  echo bar
)
EOF

# or
cat <<<"$(echo foo;sleep 3;echo bar)"

Internally bash will create a safe temporary file, put the output of the commands into it, then pass it to cat stdin. The file is destroyed as soon as the command (here cat) release it stdin.

Note usually it is better to use a stream.

Upvotes: 1

Diego Torres Milano
Diego Torres Milano

Reputation: 69208

The simplest way would be

{
 echo "foo";
 sleep 1;
 echo "bar";
} > /dev/shm/file.$$; cat /dev/shm/file.$$

Upvotes: 1

edi9999
edi9999

Reputation: 20554

One way to do this would be to use the printf command :

For example 

{
echo "foo"
sleep 1
echo "bar"
} | printf "%s\n" "$(cat)"

Would do the trick

Upvotes: 0

Related Questions