Date command

menator · 06-27-2006, 04:52 AM

Im trying to create a variable in my shell script that contains todays date but with the day and month padded with blanks:

Code:

date +%e/%m/%Y

This would return 27/06/2006. The day of the month is padded with blanks so that you get '1' instead of '01' in the day (e.g. 1/06/2006 instead of 01/06/2006). The problem is that there doesnt seem to be a way to do the same for the month of the year...well not according to the man page anyway.

Is there I way i can achieve this?

kassle · 06-27-2006, 05:12 AM

ok, so you want to show 1/6/2006. try the following script

Quote:

month=`date +%m`
zero=`echo $month | cut -c 1`
if [ $zero -le 0 ]; then
month=`echo $month | cut -c 2`
fi
echo `date +%e`"/$month/"`date +%Y`

menator · 06-27-2006, 05:36 AM

oh that worked perfectly - thanks for the help

jtshaw · 06-27-2006, 06:00 AM

You have other options too...

Code:

date +%_d/%m/%y

From the man page on my machine:

Code:

%e     day of month, space padded; same as %_d

<EDIT> ok.. I can read...

try:

Code:

date +%_d/%_m/%y

You can put the _ in any construct to get space padding...

Code:

j_shaw@jshawlptp:~$ date +%_d/%_m/%y
27/ 6/06