RE: Silly little regex question

["Dan Roozemond" <dan+debian@banaan.org>]

> 
> In a shell script:
> 
> `eval date +%j`
> 
> returns the so-called "Julian" date (a misnomer I believe, but there
> it is), the number of days into the present year, aka "yearday":
> 
> $ echo `eval date +%j`
> 036
> 
> I want a regex that will strip the leading zero _or_ zeros (if it's
> January) from the yearday, using the shell contructions:
> 
> ${string#substring}   # strip shortest occurence of substring from the
> front of string
> or
> ${string##substring} # strip longest occurence of substring from the
> front of string
> 

If you do not insist on using shell constructions, you can use 'sed':
$ date +%j | sed 's/^0\+\(.*\)/\1/'
36

or inside a shell script you might use:
T=`date +%j | sed 's/^0\+\(.*\)/\1/'`

If you insist on using this shell construction, it will look like:
T=`date +%j`
U=${T##"0"}
echo $U

> where substring is a regex.
> 

This sentence makes me think I misunderstood your question.

HTH anyway,
Dan