Add '\\n' after a specific number of delimiters

Question

How can I add a \n after each four ; delimiter in a CSV file (with bash)?

Input file sample:

aaaa;bbbbbb;cccc;ddddd;eeee;ffff;gggg;hhhh;iii;jjjj;kkkk;llll;

Output needed :

aaaa;bbbbbb;cccc;ddddd
eeee;ffff;gggg;hhhh
iii;jjjj;kkkk;llll

Answer 1

Using (GNU) sed:

... | sed -r 's/([^;]*;){4}/&\n/g'

[^;]*; matches a sequence of characters that are not semicolons followed by a semicolon.

(...){4} matches 4 times the expression inside the parentheses.

& in the replacement is the whole match that was found.

\n is a newline character.

The modifier g make sed replace all matches in each input line instead of just the first match per line.

Answer 2

Read each line into an array, then print 4 groups at a time with printf until the line is exhausted.

while IFS=';' read -a line; do
    printf '%s;%s;%s;%s\n' "${line[@]}"
done < input.txt

Answer 3

Perl solution:

perl -pe 's/;/++$i % 4 ? ";" : "\n"/ge; chomp'

Only works if the number of fields is divisible by four.

Answer 4

This might work for you (GNU sed):

sed 's/;/\n/4;/./P;D' file