Count number of occurrences of a pattern in a file (even on same line)
When searching for number of occurrences of a string in a file, I generally use:
grep pattern file | wc -l
However, this only finds one occurrence per line, because of the way grep works. How can I search for the number of times a string appears in a file, regardless of whether they are on the same or different lines?
Also, what if I'm searching for a regex pattern, not a simple string? How can I count those, or, even better, print each match on a new line?
To count all occurrences, use -o . Try this:
echo afoobarfoobar | grep -o foo | wc -l
And man grep of course (:
Update
Some suggest to use just grep -co foo instead of grep -o foo | wc -l grep -o foo | wc -l .
Don't.
This shortcut won't work in all cases. Man page says:
-c print a count of matching lines
Difference in these approaches is illustrated below:
1.
$ echo afoobarfoobar | grep -oc foo
1
As soon as the match is found in the line ( a{foo}barfoobar ) the searching stops. Only one line was checked and it matched, so the output is 1 . Actually -o is ignored here and you could just use grep -c instead.
2.
$ echo afoobarfoobar | grep -o foo
foo
foo
$ echo afoobarfoobar | grep -o foo | wc -l
2
Two matches are found in the line ( a{foo}bar{foo}bar ) because we explicitly asked to find every occurrence ( -o ). Every occurence is printed on a separate line, and wc -l just counts the number of lines in the output.
A belated post:
Use the search regex pattern as a Record Separator (RS) in awk
This allows your regex to span n -delimited lines (if you need it).
printf 'X n moo Xn XXn' |
awk -vRS='X[^X]*X' 'END{print (NR<2?0:NR-1)}'
Try this:
grep "string to search for" FileNameToSearch | cut -d ":" -f 4 | sort -n | uniq -c
Sample: grep "SMTP connect from unknown" maillog | cut -d ":" -f 4 | sort -n | uniq -c
6 SMTP connect from unknown [188.190.118.90]
54 SMTP connect from unknown [62.193.131.114]
3 SMTP connect from unknown [91.222.51.253]
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