I have this small script to test a FTP site:
#!/bin/bash wget -O /dev/null ftp://someftpsite:[email protected]/testdump300 2>&1 | \ grep '\([0-9.]\+ [M]B/s\)' >> wget300.log And it shows output like this:
2018-07-26 22:30:06 (22.7 MB/s) - '/dev/null' saved [104857600] Ok, and now I just want to have it like this:
2018-07-26 22:30:06 22.7 Anyone who can help? I suspect I should use awk or sed?
Using awk:
$ wget -O /dev/null ftp://someftpsite:[email protected]/testdump300 2>&1 | \ awk '/[0-9]+ [M]B\/s/{ sub("\\(",""); print $1,$2,$3 }' >> wget300.log This will eliminate your need for grep as awk can also search for regex patterns. It will remove the ( before the speed and then print columns 1, 2, and 3. (date, time, and speed).
Here's an alternative using sed:
$ wget -O /dev/null ftp://someftpsite:[email protected]/testdump300 2>&1 | \ sed 's/(//;s/ [[:alpha:]]\+\/s.*$//' >> wget300.log How it works:
s/(//; - deletes the first parenthesiss/ [[:alpha:]]\+\/s.*$// - deletes everything starting from the space + 'MB/s' to the end .*$.Another way using perl:
$ wget -O /dev/null ftp://someftpsite:[email protected]/testdump300 2>&1 | \ perl -lne 'print "$1 $2" if /^(.*)\s\((\S+)/' >> wget300.og How it works:
$1 and $2 variables. In this case we're matching everything up to but not including the space + parenthesis in $1 and everything after the open parenthesis that's not a space \S+ in the 2nd variable, $2.