Ygtjki

I need count some patterns in logs, I can use

grep aaa ./logs | wc -l

grep bbb ./logs | wc -l

is there a easy way to do all things in one line? like

cat ./logs | grep -c aaa | grep -c bbb #didn't work.

You can use the following command:

$ grep -oh -e aaa -e bbb ./logs | sort | uniq -c

From man grep, you can read:

-o, --only-matching
Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.

Also, for -h:

-h, --no-filename
Suppress the prefixing of file names on output. This is the default when there is only one file (or only standard input) to
search.

The -e is used to match either one. Then, the results are sorted and counted using uniq.

You can use the -e PATTERN flag.
In order to count how many lines contain either or both aaa or bbb:

grep -e aaa -e bbb ./logs | wc -l

If you want to count aaa and bbb separately, check the solution by Khaled.

Using awk:

awk '/aaa/ count["aaa"]++ 
 /bbb/ count["bbb"]++ 
 END for (pat in count) print count[pat], pat ' file

This would update the count associated with the matching pattern whenever that pattern matches. At the end, a list of counts and the corresponding pattern are outputted.

I Tried with below command

veen:~$ for i in aaa bbb; do perl -pne "s/n/ /g" i.txt| awk -v i="$i" 'print gsub(i,$0)';echo $i; done| sed "N;s/n/ /g"| awk 'print $2,$1'

output

veen:~$ for i in aaa bbb; do perl -pne "s/n/ /g" i.txt| awk -v i="$i" 'print gsub(i,$0)';echo $i; done| sed "N;s/n/ /g"| awk 'print $2,$1'

aaa 1
bbb 2