mjhjmtu

I want to count the number of 6-letter words that start with 'bar' and end with 'i' or 'n'.

For example, given this file:

barxxi
barxxc
barxxn
barqwq
barovo

the output should be 2 (integer), since these lines match:

barxxi
barxxn

My code is the following, but it seems to be wrong:

grep -c 'bar??[ni]' /path/file.txt

Use grep -o which outputs only the matched text, multiple time per line if necessary.

grep -o '<bar[^ ][^ ][ni]>' /path/file.txt | wc -w

The < and > match the beginning and end of a word.
The [^ ] matches a non-space character.

Replace your question marks ? with periods . and your grep should work, as long as your input is only one word per line.

grep -c 'bar..[ni]' /path/file.txt

awk ' $i ~ /[[:space:]]bar.*[in]/) count++ END print count ' filename

The above awk solution should also work. Taking account of multiple entries on each line, awk looks from the first space delimited entry in the file on each line to the last entry (NF) if then pattern matches against regular expressions with ~ and increments a count accordingly. Two if statements are created to take account or a beginning of line (^) and bar as well a space and bar. At the end, the count is printed to screen.