How can I get the hostname from a file using awk and regex or substring

Question

The file name is in a format like this:

YYYY-MM-DD_hostname_something.log

I want to get the hostname from the filename. The hostname can be any length, but always has a _ before and after. This is my current awk statement. It worked fine until the hostname length changed. Now I can't use it anymore.

awk 'BEGIN { OFS = "," } FNR == 1 { d = substr(FILENAME, 1, 10) } { h = substr(FILENAME, 12, 10) } $2 ~ /^[AP]M$/ && $3 != "CPU" { print d, $1 "" $2, h, $4+$5, $6, $7+$8+$9}' *_something.log > myfile.log

Cyrus · Accepted Answer · 2015-08-16 16:55:40Z

2

echo 'YYYY-MM-DD_hostname_something.log' | awk -F"_" '{print $2}'

Output:

hostname

I suppose your hostname contains no _.

answered Aug 16, 2015 at 16:55

Cyrus

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Cyrus Over a year ago

Try this: find -name "*_something.log" | awk -F"_" '{print $2}'

Jens · Accepted Answer · 2015-08-17 14:47:49Z

1

$ ls YYYY-MM-DD_hostname_something.log | cut -d _ -f 2
hostname

The cut(1) utility is POSIX and accepts the -d _ option to specify a delimiter and -f 2 to specify the second field. It has got a few more nifty options that you can read about in its fine manual page.

edited Aug 17, 2015 at 14:47

answered Aug 16, 2015 at 16:56

Jens

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Comments

Mahesh Kharvi · Accepted Answer · 2015-08-17 06:06:48Z

0

Since you have mentioned you need to modify your awk code, replace your substr function with split.

split(FILENAME,a,"_");date = a[1];host = a[2]

split the FILENAME value into array a with _ as FS.
a[1] will contain date
a[2] will contain the hostname value.

edited Aug 17, 2015 at 6:06

answered Aug 17, 2015 at 5:11

Mahesh Kharvi

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