how to find lines that contain consecutive vowels
$ (filename) | sed \'/[a*e*i*o*u]/!d\'
To find lines that contain consecutive vowels you should consider using
sed -n '/[aeiou]\{2,\}/p' file
Here, [aeiou]\{2,\}
pattern matches 2 or more occurrences (\{2,\}
is an interval quantifier with the minimum occurrence number set to 2
) and [aeiou]
is a bracket expression matching any char defined in it.
The -n
suppresses output, and the p
command prints specific lines only (that is, -n
with p
only outputs the lines that match your pattern).
Or, you may get the same functionality with grep
:
grep '[aeiou]\{2,\}' file
grep -E '[aeiou]{2,}' file
Here is an online demo:
s="My boomerang
Text here
Koala there"
sed -n '/[aeiou]\{2,\}/p' <<< "$s"
Output:
My boomerang
Koala there