How to use sed to replace every occurrence of a pattern minus one case?

Question

I have a temporary file generated by a cron job every half hour that generates output like this:

---- kind=<data> field=<value> humankind=<data> kind=<data> field=<value> humankind=<data> kind=<data> field=<value> humankind=<data>

I want to format it so it looks like this:

---- kind=<data> field=<value> humankind=<data>
kind=<data> field=<value> humankind=<data>
kind=<data> field=<value> humankind=<data>

I have tried the following sed expressions:

sed -r 's|\s(kind=)|\n\1|g' /path/to/file

But then the output looks like:

----                                         <------ The first line should be here
kind=<data> field=<value> humankind=<data>
kind=<data> field=<value> humankind=<data>
kind=<data> field=<value> humankind=<data>

Using this expression:

sed -r 's|[^-]{4} (kind=)|\n\1|g' /path/to/file

Generates the output I want, but with weird behavior:

---- kind=<data> field=<value> humankind=<incomplete data>
kind=<data> field=<value> humankind=<incomplete data>
kind=<data> field=<value> humankind=<incomplete data>

For some reason, the last field (which contains part of the pattern used in the sed expression) is only printing the first two characters of 'incomplete data'.

What am I doing wrong?

Answer 1

The problem with sed -r 's|\s(kind=)|\n\1|g' /path/to/file is that you insert a new line for all ' kind' occurrences, including the first, so you get an extra newline after ----. Instead, try

sed -E 's|\s(kind=)|\n\1|2g' /path/to/file

as it will skip the first match.

The second one fails because it matches 4 chars before ' kind', and that is part of your data (which must be 6 chars long)!