Question

I need to get the browser version which is followed by browser name and / in my log file. Example

Mozilla/5.0 (Windows NT 6.1; WOW64; rv:17.0) Gecko/20100101 Firefox/17.0.5.7

Here i would like to extract the version which comes after Firefox i.e (17.0.5.7). I have tried the following, but no success. As i am using AIX, i am unable to make use of -O flag.

grep -w 'Firefox\/[^ ,]\+' $line
sed 's/^.*\(Firefox\/[^ ,]*\).*$/\1/' $line

Also i need use the command inside while do loop like below

while read -r line
do
case $line in    
*Firefox*) print "<browser Firefox>\"" command_to_be_executed $line "\"</browser>";;
*) print "<browser>\"Unknown\"</browser>";;
esac
done < filename

Thanks in adavance.

Answer 1

Since it’s the last field in the line, you can do:

awk -F'[ /]' '$(NF-1)=="Firefox"{print $NF}' file.log

It prints the lines which checks whether the it is “Firefox” and prints the version followed.

---

Simply do echo on line and get the version.

while read -r line
do
case $line in    
*Firefox*) version=$(echo $line | awk -F'[ /]' '{print $NF}' )
print "<browser Firefox>\"" $version "\"</browser>";;
*) print "<browser>\"Unknown\"</browser>";;
esac
done < filename