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# 1  
Old 11-17-2009
Date format
Hi,

I've got a UNIX (Solaris 9 ksh) command to extract from a log file the informations I need. I am searching a way to format the date (YYYY/MM/DD to DD/MM/YYYY) . From the log file, the date is a unique string so I am good enought to make it right. Could you please help me ?

Here is my command :

Code:
cat file_name | grep "COMMAND PROCESS FINISHED IN" | awk -F" " '{print $10" "$6" "$12" "$13" "$1" "$2}' | cut -d "," -f1 | sort -k 1 | sed 's/-/\//g'

here is what i've got :

Code:
COMMAND 259 FINISHED IN 2009/10/08 11:38:30

here is what am looking for :

Code:
COMMAND 259 FINISHED IN 08/10/2009 11:38:30

Thanks in advanced for your help
# 2  
Old 11-17-2009
Quote:
Originally Posted by Aswex
...
here is what i've got :

Code:
COMMAND 259 FINISHED IN 2009/10/08 11:38:30

here is what am looking for :

Code:
COMMAND 259 FINISHED IN 08/10/2009 11:38:30

...
One way to do it with Perl:

Code:
$
$ cat f9
COMMAND 259 FINISHED IN 2009/10/08 11:38:30
$
$ perl -lne '/(.*? )(\d+)\/(\d+)\/(\d+)( .*)/ && print "$1$4\/$3\/$2$5"' f9
COMMAND 259 FINISHED IN 08/10/2009 11:38:30
$
$

What's the output of the following ?

Code:
grep "COMMAND PROCESS FINISHED IN" file_name

tyler_durden
# 3  
Old 11-17-2009
Hello durden_tyler,

is there a way to do it without PERL ?

about :

Code:
What's the output of the following ?

 grep "COMMAND PROCESS FINISHED IN" file_name

What do you mean ?

Thanks for your help.
# 4  
Old 11-17-2009
Quote:
Originally Posted by Aswex
Hello durden_tyler,

is there a way to do it without PERL ?

about :

Code:
What's the output of the following ?

 grep "COMMAND PROCESS FINISHED IN" file_name

What do you mean ?

Thanks for your help.
We need to know the format on the line(s) containing the serached string "COMMAND PROCESS FINISHED IN".
Please post the output of the grep command.

Jean-Pierre.
# 5  
Old 11-17-2009
Quote:
Originally Posted by Aswex
...
is there a way to do it without PERL ?
...
Sure there is. What language do you prefer ?

tyler_durden
# 6  
Old 11-17-2009
Quote:
Originally Posted by Aswex
I've got a UNIX (Solaris 9 ksh) command to extract from a log file the informations I need. I am searching a way to format the date (YYYY/MM/DD to DD/MM/YYYY) . From the log file, the date is a unique string so I am good enought to make it right. Could you please help me ?

Here is my command :

Code:
cat file_name | grep "COMMAND PROCESS FINISHED IN" | awk -F" " '{print $10" "$6" "$12" "$13" "$1" "$2}' | cut -d "," -f1 | sort -k 1 | sed 's/-/\//g'

here is what i've got :

Code:
COMMAND 259 FINISHED IN 2009/10/08 11:38:30

here is what am looking for :

Code:
COMMAND 259 FINISHED IN 08/10/2009 11:38:30


I don't understand why you would want to change the date from the ISO standard to an ambiguous format.

Code:
## untested
nawk '
/COMMAND PROCESS FINISHED IN/ {
   $0 = $10" "$6" "$12" "$13" "$1" "$2
   sub( /,.*/,"" )
   split( $5,a, "-")
   $5 = a[3]"/"a[2]"/"a[1]
   print
}' file_name

# 7  
Old 11-18-2009
Dear aigles, durden_tyler and cfajohnson,

sorry to be so late to respond.

Code:
We need to know the format on the line(s) containing the serached string "COMMAND PROCESS FINISHED IN".
Please post the output of the grep command.

Jean-Pierre.

here is the grep command output :

Code:
2007-10-30 14:50:05,544 [Command Thread-5] Command 145 INFO  saturne.service.command.manager.ExecutableCommand - COMMAND PROCESS FINISHED IN 11448 ms

Code:
Sure there is. What language do you prefer ?

tyler_durden

I am supposed to used the shell (ksh).

Code:
 don't understand why you would want to change the date from the ISO standard to an ambiguous format. cfajohnson

dear cfajohnson,

I have to change this to make the date format compliant to the reporting format used by the administrator's. I can't deal with them to change this.

Thanks to all, you help me so much.
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