[plug] Grep not following my regex
tim.bowden at westnet.com.au
Tue Sep 15 10:36:50 WST 2009
On Tue, 2009-09-15 at 09:38 +0800, Tim Bowden wrote:
> On Tue, 2009-09-15 at 09:11 +0800, Tim wrote:
> > I have been doing some string extraction for an application I've
> > written. I needed to extract a number (with decimal point) from a one
> > line string. I had the grep working, but then having moved to another
> > computer, it stopped working. I did some debugging on the computer I'd
> > moved to, to see why my application no longer worked, and narrowed it
> > down to this grep.
> > egrep -o '[0-9.]*'
> > By changing the grep to the following, I managed to get it working again.
> > egrep -o '[0-9.]+'
> > Now I know that the + may make more sense now, by forcing a match, but
> > I can't see why the first regex stopped working.
> > The string it's matching against is:
> > You currently have 347.454MB remaining
> > and obviously it just pulls out the 347.454.
> > Other than bash stealing the * from the grep (which I am sure
> > shouldn't be happening due to the quotes), can someone let me know why
> > the first regex stopped working? (Also, I was moving from an Ubuntu
> > system to a Fedora system)
> > Thanks
> > Tim
> The '.' has special meaning in a regex. Escape it to make it just a
> Without looking at your problem in detail, here is a possible regex for
> a decimal point number.
> grep -o [0-9]+\.[0-9]+
Meh. What a load of horse shit. \ doesn't escape the '.' at all; Perl
it ain't. And it should be egrep. Happens to work anyway, so long as
your strings are as you say (otherwise it might not be quite what you're
<hangs head in shame>
More information about the plug