Replace a word with multiple lines using sed?

Question

I\'m working on a bash-script that has to prepare an E-Mail for being sent to a user.

It aggregates some data, which ends up being multiple lines of stuff. For the e

Answer 1

I would suggest simply replacing sed with perl command like this:

perl -i.bak -pe 's/_data_/$ENV{"DATA"}/g' mail.tpl 

Answer 2

You can do this with AWK using variable substitution. We can set a variable in AWK using -v, and then use AWK's gsub function to substitute all occurrences of a regular expression with that variable.

For example, if the file test has the following contents ...

foo
bar
blah _data_and_data_
foo
_data_ foobar _data_ again

... and the Bash variable $DATA is ...

1
2
3
4
5

... then awk -v r=$DATA '{gsub(/_data_/,r)}1' test replaces all occurrences of the regular expression _data_ in the file test with the contents of $DATA, resulting in the following:

foo
bar
blah 1
2
3
4
5and1
2
3
4
5
foo
1
2
3
4
5 foobar 1
2
3
4
5 again

Answer 3

ring bearer's answer didn't work for me; I think the usage of tr there is wrong, and the way it's written, it simply strips away newlines by use of echo.

Instead, I used sed. I used code from another answer to replace newlines (credit: Zsolt Botykai). I also expected some dollar signs ($) in my input so I took care of that too. You might need to add other input handling. Note the use of double quotes in echo to preserve newlines.

DATA="$(cat whatever)"
ESCAPED_DATA="$(echo "${DATA}" | sed ':a;N;$!ba;s/\n/\\n/g' | sed 's/\$/\\$/g')"

Then you can use ${ESCAPED_DATA} in sed:

cat input | sed 's/one liner/'"${ESCAPED_DATA}"'/' > output 

Just thought I'd share.