--- Craig White <craigwhite@azapple.com> wrote:
> On Mon, 2004-03-22 at 17:45, Bill Jonas wrote:
> > On Mon, Mar 22, 2004 at 05:29:35PM -0700, Craig
> White wrote:
> > > I will be dammed but I can't make the line above
> this one work...this is
> > > where the problem is...
> > > i=`echo $line | cut -f1`;
> > 
> > Try:
> > i=`echo $line | cut -d' ' -f1`;
> > instead.
> ----
> yup that did it
> 
> this is too hard for me ;-)
> 
> Craig
> \

The part... -d' ' -f1 set the delimiter to the first
field.. per my unix1 class just a couple hours ago!
The space between the two ' ' sets the delimiter to
the space bar or white space in the output of $line
(if I understood Gerald, my instructor correctly.

Excerpt from the 'class' Here, we needed the 
':' or colon delimiter, but you need the ' ' white
space delimiter

 $ cut -f1 -d":" /etc/passwd | and on to another stdin


      The 'cut' command prints the values of field
      one found in the /etc/passwd file having colon
      delimited fields. 

He wrote one in the class as an example and used the
white space delimiter.

Hope that explains that/helped!!!

=====
Cyclists should expect and demand safe 
accommodation on our public roads, 
just as does every other user.  

Nothing more is expected. 

Nothing less is acceptable!

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