Shell Programming and Scripting

Hi
I am dealing with the following string:

Date: Thur, 13 March 2011 01:01:10 +0000

I asked for help in another topic that converted a similar string:

Date: Thur, 13 March 2011 9:50 AM

To a 24 hr standard. The problem is that it comes out as:

Date: Thur, 13 March 2011 9:50:00 +0000

I need to have it so that the first number in the time stamp ( in this case 9) to appear as 09:50:00.

Basicly if the first set of number in the time stamp is only 1 character long then append a 0 at the start.

This is the code used to convert the 12hr clock to 24hr:

Code:

awk '/PM/{split($6,_1,":");$6=_1[1]+12":"_1[2]}/AM|PM/{$6=$6":00";$7=$8;NF=7}1' infile

I thought I could fix it by replacing

Code:

{$6=$6

with

Code:

{$6=0$6

but no go as it appends a 0 at the start even when the time is PM

Thanks

Try:

Code:

perl -pe 's/\b\d:\d\d:\d\d\b/0$&/g' file

thank you!
I havent really touch much of perl, what is it that it is doing?

It is matching any string in form of "*:**:**", where stars are digits, and it is putting 0 in front of it. \b (word border) at the beginning of the match makes sure that only single digit number in first position is matched.

or you can try wıth sed

Code:

# cat file
Date: Thur, 13 March 2011 9:50:00 +0000

Code:

# sed 's/\(.*\)\(\b[0-9]\)\(:[^ ]*:[^ ]*.*\)/\10\2\3/' file
Date: Thur, 13 March 2011 09:50:00 +0000

it is find your part of string within 9:50:00 if it has single char in "9:.*:.*"
and append a zero to beginning its.

\(.*\) portion of full string so that `Date: Thur, 13 March `
\(\b[0-9]\) portion of full string so that `9`
\(:[^ ]*:[^ ]*.*\) portion of full string so that `50:00 +0000`

regards
ygemici