Bashscripts.org

In a script I need to put the date in variables. So I've used the following:

month=`date '+%m'`
day=`date '+%d'`
year=`date '+%Y'`

which for today's date $day=08

I then use the values to grep lines from a file. The problem is the file I've trying to pull from, does not use leading zeros, so june 8th 2005 is recorded as 6-8-2005 not 06-08-2005, which is what the date commands output would give me. I can't see anyway to strip the leading zeros. I did a cheap trick of:

let "--day" ; let "day++"

which stripped the leading zero, but It fails on the 8th. because the bash is assuming its an octal number because of leading zero.

I thought of using cut, but that would fail in the case of 10, 20 or 30.

Is there some easy way to do this?

Try:


%e day of month, blank padded ( 1..31)

or you could even try



ofcourse crouses way would be easiest...but i just LOVE string manipulation

-e works on the day, but not on the month, I'm shocked I missed that in the man page though.

I figured out your suggestion from another post.

Is there anything like that works on a chars in a single line, like grep works on lines in a file?

maybe explain a little more of what you mean...
do you mean something that can search for particular strings within a string or ??

Yes, something that would let you search a string for a substring.

Or you could say something like find this substring, in a string, then return all text after it, or your can specify a number of chars before or after the substring to return:

But I guess that could be done with cut now that I think about it.