i would like to know what's the best way to grep lines from text which have more than specific number of special characters.
Let's say that you already know that each line have 4 comma , and you would like to grep lines which have more than 4 comma ,
Example
hi,hello,how,are,you catch,me,then,say,hello,then Output
catch,me,then,say,hello,then Perl solution:
perl -ne 'print if tr/,// > 4' -n reads the file line by lineTo print the lines with less than 4, just change > to <.
Using the grep command:
grep -E '(,.*){5}' myfile does the job. Explanation:
-E: use an Extended Regex...
'(,.*): ... to find one comma followed by any number of characters, even zero...
{5}': ... and repeat the previous pattern 5 times.
If you want to grep lines with less than 4 commas, you'd need:
grep -xE '([^,]*,){0,3}[^,]*' myfile This time, we need -x so the pattern is anchored at both start and end of the line so it matches the full line. And we use [^,]* instead of .* as the latter would otherwise happily match strings containing ,s as . matches any character.
Another approach is to reverse with -v the previous approach. "Fewer that 4" is the same as not "at least 4", so:
grep -vE '(,.*){4}' myfile The awk version:
awk -F, 'NF > 5' myfile Achieved result by below one liner
l=`awk 'BEGIN{print }{print gsub(",","")}' example.txt |sed '/^$/d' |awk '$1 > 4 {print NR}'`;sed -n ''$l'p' example.txt output catch,me,then,say,hello,then 
grep -E '(.*,){5}' file?grep -E '^([^,]*,){,3}[^,]*$' file? This might help: The Stack Overflow Regular Expressions FAQ