Match only the first paragraph using bash

This is a tailor-made problem for gnu awk by using a custom record separator. We can use a custom RS that breaks file data by 2 or more of an optional \r followed by \n:

awk -v RS='(\r?\n){2,}' 'NR == 1' file

This outputs:

Foo
* Bar

If you want awk to be more efficient when input is very big:

awk -v RS='(\r?\n){2,}' '{print; exit}' file

You can use a GNU grep like this:

grep -Poz '(?s)^.+?(?=\R{2}|$)' file

See the PCRE regex demo.

Details

• (?s) - a DOTALL inline modifier that makes . match all chars including linebreak chars
• ^ - start of the whole string
• .+? - any 1 or more chars, as few as possible
• (?=\R{2}|$) - a positive lookahead that matches a location immediately followed with a double line break sequence (\R{2}) or end of string ($).

For GNU awk if the paragraphs are separated by \r\n\r\n or \n\n:

$ awk -v RS="\r?\n\r?\n" '{print $0;exit}' file

Output:

Foo
* Bar