Grep out only today's date

Shell Programming and Scripting
1

Hi,

I have some log files. I've been asked to grep out error messages that have happened ONLY today. (However, the logs keep messages a long time so they have error messages going back weeks)

They log details in these fields (order):

Month Day Time Server message

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:

cat 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?

Also, i would then like to email out if the date matches today's date....and if not, email out that nothing has changed or been logged today.

2

Please do not leave people guessing. Show a representative sample of input, desired output, attempts at a solution and specify what OS and versions being used, or this thread will be closed.

That's also a useless use of cat award.

3

excuse my ignorance, i'm a newbie to Unix.

i'm using Solaris

the output of the log files are:

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:

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

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

<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?

Also, i would then like to email out if the date matches today's date....and if not, email out that nothing has changed or been logged today.

thanks for any responses

4

How about..

$ mon=$(date +'%b')
$ day=$(date +'%d')
$ grep -E "$mon[ ]+$day" xx
May  23 20:34:33 server_name Process: Error_message_timed_out
May 23 20:34:33 server_name Process: Error_message_timed_out
May    23 20:34:33 server_name Process: Error_message_timed_out
$ 
$ 
$ 
$ cat xx
May  23 20:34:33 server_name Process: Error_message_timed_out
May 23 20:34:33 server_name Process: Error_message_timed_out
other msg
other msg
May    23 20:34:33 server_name Process: Error_message_timed_out
May 19 20:34:33 server_name Process: Error_message_timed_out
other msg
May 19 20:34:33 server_name Process: Error_message_timed_out
May 19 20:34:33 server_name Process: Error_message_timed_out
May 19 20:34:33 server_name Process: Error_message_timed_out
$

You have to deal with the other patterns if any. Just update the pattern.

1 Like
6

hi thanks for your help.

that seemed to do the trick! thanks very much!

7

first put the contents of file in a temp file with trimming the space then try to search the pattern, words will be separated by max 1 space.

cat file_name|tr -s " " >temp_file
1 Like
8

Please also post sample data for a day of the month with a single-digit day.
Also sample data for every possible variation in the date field of the input record format.

9

Almost, you need to leave out the "|"

<file_name cut -d " " -f 1,2

or

cut -d " " -f 1,2 < filename

To be independent of spacing it is best to use awk:

awk '{print $1,$2}' < filename
1 Like
10

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.

or

Here you can print any desired column like $1,$2,$3,etc. Just ensure u put a comma after column number (like $1).

1 Like