Print word that doesn't contain character

Question

My file is in the following format:

this!,is!,another!,test,yes!
this!,is!,another!,column,yes!
no,not!,another!,column

My output should be:

test
column
no

It shouldn't contain the '!' character.

I've tried multiple sed and (e)grep commands but none of them are quite working.

Answer 1

Try something like this:

tr ',' '\n' < file.txt | grep -v \!

Answer 2

Assuming that you really meant...

test
column
no
column

...here's an awk solution:

awk -v RS=, '{ for (i=1; i<=NF; i++) if($i !~ /!/) print $i; }'

Answer 3

requires GNU grep:

grep -oP '(?<=^|,)[^!]+(?=,|$)'

based on your input, that reports:

test
column
no
column

If you don't want "column" to appear twice: grep -oP '(?<=^|,)[^!]+(?=,|$)' | sort -u

Answer 4

Perl is made for these kind of tasks. How about this perl (v5.10 and above) one liner:

perl -aF, -nE 'for (@F) { say $_ unless m/!/ }'

• -a for autosplit mode
• -F, split input lines on comma
• -n automatically insert code to iterate over input lines, do not automatically print
• -E execute supplied per code in the loop

If you have an older perl, just do this: s/-E/-e/, s/say/print/

Answer 5

assuming your text file is test.txt, the following:

for word in $(cat test.txt | sed -e 's/,/ /g'); do if [ ! "$( echo "$word" | grep '\!' )" == "$word" ]; then echo "$word"; fi; done

gives me:

test
column
no
column

Answer 6

also there is solution in pure sed

sed 's/\([,]\)*[^,]*![^,]*[,]*/\1/g;s/,$//;s/,/\n/g' file.txt