Shell Programming and Scripting

I have a file that I need to parse using a script. The dates in the file are displayed in the format:

Code:

Mar 2, 2011 9:09:31 PM

I have tried using the date command %e and %l but it pads an extra space for the day and hour if they are single digits.
So this I used a normal date command:

Code:

date=`date '+%b %d, %Y %r'`

and I'm trying to remove the 0s at position 5 and 14 (eg: Mar 03, 2011 11:38:35 AM)
Is there someway that I can do this. I tried the sed command

Code:

sed -e s/.//5 -e s/.//13

but it will remove anything at position 5 and 14. I want it to remove the value only if it is 0 (zero).

Any help is greatly appreciated.

Code:

sed 's/^\(....\)0/\1/' input

could also be written

Code:

sed 's/^\(.\{4\}\)0/\1/' input

Code:

sed 's/^\(.\{13\}\)0/\1/' input

My advice : Take care : since you remove the 0 instead of replacing it with a <space> character you should FIRST remove the 0 that is the most on the RIGHT

Consider the following example :

Code:

Mar 03, 2011 10:38:35 AM

the 0 of the 10 is currently at the 15th position :

if you remove the 0 on fifth position, all character will be shifted to left so that the 0 of the 10 will be on the 14th position

Look the difference :

Code:

# echo "Mar 03, 2011 10:38:35 AM" | sed 's/^\(....\)0/\1/;s/^\(.\{13\}\)0/\1/'
Mar 3, 2011 1:38:35 AM

Code:

# echo "Mar 03, 2011 10:38:35 AM" | sed 's/^\(.\{13\}\)0/\1/;s/^\(....\)0/\1/'

Mar 3, 2011 10:38:35 AM

Another solution could be to replace the unexpected 0 with a space character, so the characters will keep their positions.

HOW ABOUT THIS?

Code:

date '+%b %d, %Y %r' |awk -F"[ ,:]" '{$2+=0;$6+=0;print $1" " $2", "$3" "$4" "$5":"$6":"$7" "$8}'

Thanks, this works like a charm.
I really didn't think about the part where we will have to remove the right 0 first, and then remove the left one.
Sometimes you just miss the obvious.

Thanks a whole bunch.

Alternatively, since in this format a leading zero is always preceded by a space, the following should also work:

Code:

sed 's/ 0/ /g'

Regards,
Alister

wow! plain and simple. And to think i was complicating this so much.

can be fixed directly by sed:

Code:

your_date_command |sed 's/  */ /g'