1

I am trying to calculate what is the % of successful queries in apache log. I have two commands:

cat access_log|cut -d' ' -f10|grep "2.."|wc -l

and

cat access_log|cut -d' ' -f10|wc -l

They return me the number of successful queries and total queries number. I want to calculate what is the % of successful requests using bash and if it is possible - it should be 1 line script. It suppose to output just the % number like - 50 or 12 without any additional info.

I tried to use bc with it but failed because of lack of knowledges. Can somebody help me?

1

Try this:

echo $(( 100 * $( cut -d' ' -f10 access_log|grep "2.."|wc -l) / $(cut -d' ' -f10 access_log|wc -l) ))

Bash can only handle integers.

0

Using awk and only iterates through the logfile once:

awk '{if ((199 < $9) && ($9 < 300)) {SUMOK++} else {OTHER++}} END { printf "%d\n", ((SUMOK/NR)*100)}' access_log
0

You can try this. Replace $9 with correct field number of status code.

awk '{if ($9 == 200)  no_of_200+=1 } END{ perc=(no_of_200/NR)*100; print perc}' access.log
0
sed -ne'\|^\([^ ]*  *\)\{9\}2..|=;$=;$s|.*|2ksmzlm/p|'|dc

...works...

for example:

printf %s\\n 1 2 3 4 5 6 7 8 9 10 |
sed -ne'/1/=;$=;$s|.*|2ksmzlm/p|p'|dc

.20

...to show that sed matched the 1 pattern against 20% of its input lines.

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