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Grep out only today's date

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# 8  
Old 05-25-2012
Quote:
Originally Posted by horhif
[..]excuse my ignorance, i'm a newbie to Unix.

i'm using Solaris

the output of the log files are:

Code:
May 19 20:34:33 server_name  Process: Error_message_timed_out

I can grep out the month and day. However, on some of the log files....there is 1 space between Month and Day, and on other log files there is 2 spaces.

SO my grep works with the 1 space, but not the 2 space log files.

this is my command:

Code:
cat file_name | cut -d " " -f 1,2

tho i'm guessing (after getting my award) Smilie that it should be

Code:
<file_name | cut -d " " -f 1,2

how do i get it to grep out the dates even with 2 spaces between the Month and Time?
Almost, you need to leave out the "|"
Code:
<file_name cut -d " " -f 1,2

or
Code:
cut -d " " -f 1,2 < filename

To be independent of spacing it is best to use awk:
Code:
awk '{print $1,$2}' < filename

This User Gave Thanks to Scrutinizer For This Post:
horhif
# 9  
Old 05-25-2012
Since your data is stored in log files as records, using 'awk' would be good option. AWK deals with column-oriented text data. Try this out.
Quote:
cat <file name> | awk '/May 19/ { print $0 }'
or
Quote:
awk '/May 19/ { print $0 }' < <file_name>
Here you can print any desired column like $1,$2,$3,etc. Just ensure u put a comma after column number (like $1).
This User Gave Thanks to sam_bd For This Post:
horhif
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