Shell Programming and Scripting

Hi All,

I want to read the log file for last 30mins logs with time stamps.

Am using below command but, it is not working for me

Code:

awk -F - -vDT="$(date --date="30 minutes ago" "+%b %_d %H:%M:%S")" ' DT < $1' log.file >tmp.txt

log file time format is 2016-09-27 14:00:25,192

Moderator's Comments:

Use code tags, thanks.

So you have a log file with a date an time format as produced by the date command:

Code:

date "+%Y-%m-%d %H:%M:%S"

which give you a string similar to:

Code:

2016-09-28 05:19:09

and you extract the date field from this string to be just the year by using

Code:

awk -F- '$1'

and then you compare that to a string produced by the date command:

Code:

date "+%b %_d %H:%M:%S"

which gives you output similar to:

Code:

Sep 28 05:19:09

which means that your awk command:

Code:

awk -F - -vDT="$(date --date="30 minutes ago" "+%b %_d %H:%M:%S")" ' DT < $1' log.file

boils down to something like:

Code:

awk -F - -vDT="Sep 28 04:49:09" ' DT < $1 log.file'

which will print every line in log.file for which the expression:

Code:

"Sep 28 04:49:09" < 2016

evaluates to true (assuming that the date field in log.file starts in column 1 in each line).

Without knowing the format of log.file (specifically where the date appears on each line in that file and what field separator is used to separate the dates in that file from other text in that file, we can't help you solve your general problem. But we do know that you need to use the same format for the strings you are comparing and that you need to compare the same number of fields (i.e., year, month, day, hour, minute, and second need to be compared to year, month, day, hour, minute, and second). Comparing abbreviated month name, day, hour, minute, and second to year cannot possibly select an appropriate 30 minute period.