How can I use a file in a command and redirect out

Question

Basically I want to take as input text from a file, remove a line from that file, and send the output back to the same file. Something along these lines if that makes it any clearer.

grep -v 'seg[0-9]\{1,\}\.[0-9]\{1\}' file_name > file_name

however, when I do this I end up with a blank file. Any thoughts?

Answer 1

I usually use the tee program to do this:

grep -v 'seg[0-9]\{1,\}\.[0-9]\{1\}' file_name | tee file_name

It creates and removes a tempfile by itself.

Answer 2

You cannot do that because bash processes the redirections first, then executes the command. So by the time grep looks at file_name, it is already empty. You can use a temporary file though.

#!/bin/sh
tmpfile=$(mktemp)
grep -v 'seg[0-9]\{1,\}\.[0-9]\{1\}' file_name > ${tmpfile}
cat ${tmpfile} > file_name
rm -f ${tmpfile}

like that, consider using mktemp to create the tmpfile but note that it's not POSIX.

Answer 3

try this simple one

grep -v 'seg[0-9]\{1,\}\.[0-9]\{1\}' file_name | tee file_name

Your file will not be blank this time :) and your output is also printed to your terminal.

Answer 4

You can do that using process-substitution.

It's a bit of a hack though as bash opens all pipes asynchronously and we have to work around that using sleep so YMMV.

In your example:

grep -v 'seg[0-9]\{1,\}\.[0-9]\{1\}' file_name > >(sleep 1 && cat > file_name)

• >(sleep 1 && cat > file_name) creates a temporary file that receives the output from grep
• sleep 1 delays for a second to give grep time to parse the input file
• finally cat > file_name writes the output

Answer 5

Use sponge for this kind of tasks. Its part of moreutils.

Try this command:

 grep -v 'seg[0-9]\{1,\}\.[0-9]\{1\}' file_name | sponge file_name

Answer 6

Use sed instead:

sed -i '/seg[0-9]\{1,\}\.[0-9]\{1\}/d' file_name