split long line on a delimiter

Question

What command can I use to split input like this:

foo:bar:baz:quux

into this?

foo
bar
baz
quux

I'm trying to figure out the cut command but it seems to only work with fixed amounts of input, like "first 1000 characters" or "first 7 fields". I need to work with arbitrarily long input.

Answer 1

There are a few options:

• tr : \\n
• sed 's/:/\n/g' (with GNU sed)
• awk '{ gsub(":", "\n") } 1'

You can also do this in pure bash:

while IFS=: read -ra line; do
    printf '%s\n' "${line[@]}"
done

Answer 2

$ line=foo:bar:baz:quux
$ words=$(IFS=:; set -- $line; printf "%s\n" "$@")
$ echo "$words"
foo
bar
baz
quux

Answer 3

If your grep supports -o you can do it like this:

grep -o '[^:]\+'

Or with awk, setting the record separator to ::

awk -v RS=: 1

Or with GNU cut:

cut -d: --output-delimiter=$'\n' -f1-

Edit

As noted by Chris below, this will leave a trailing newline, this can be avoided if your awk supports specifying RS as a regular expression (tested with GNU awk):

awk -v RS='[:\n]' 1

Answer 4

don't know why ppl hate on xargs

$ xargs --version | head -1
xargs (GNU findutils) 4.7.0
...
$ printf "foo:bar:baz:quux" | xargs -d: -n1
foo
bar
baz
quux

Answer 5

In some strings I had problem with solutions above. But this worked for me:

echo $string | sed 's/\\n/ /g' | tr " " \\n