OK, got something working you might be able to use.
Just to make things more complicated:
s = '19.0 / 482.600 mm / 19.060 / 482.600 mm'
s.gsub!(/(\.0?[^0])?0+/, '\1').gsub!(/\.[\s\n]/, '')
Pepe
On Feb 28, 10:13 am, pepe <p...@betterrpg.com> wrote:
> Sorry, I didn't read your first posting fully. My solution will not
> work for the case of 482.600.
>
> Pepe
>
> On Feb 28, 10:12 am, pepe <p...@betterrpg.com> wrote:
>
> > string = '10.0'
> > string.sub!(/\.\d+/, '')
>
> > This will replace in place (sub!) any dot (\.) followed by at least
> > one number (\d+) with nothing ('').
>
> > Pepe
>
> > On Feb 27, 4:37 pm, northband_101 <northb...@gmail.com> wrote:
>
> > > Awesome - this is a start - I'll take it from here.
>
> > > Thanks!
>
> > > On Feb 27, 4:14 pm, Rob Biedenharn <r...@agileconsultingllc.com>
> > > wrote:
>
> > > > On Feb 27, 2009, at 2:47 PM, northband wrote:
>
> > > > > Hi -
>
> > > > > I would like to use gsub() to strip decimals with trailing zeros from
> > > > > a string. My string looks like this:
> > > > > --
> > > > > 19.0 " / 482.600 mm
> > > > > --
>
> > > > > I would like to end up with this:
> > > > > --
> > > > > 19 " / 482.6 mm
> > > > > --
>
> > > > > Anyone have a regular expression that can do this?
>
> > > > > Thanks!
>
> > > > It depends on how you make the string.
>
> > > > "19.0".sub(/\.?0+\z/,'') #=> "19"
> > > > "482.600".sub(/\.?0+\z/,'') #=> "482.6"
>
> > > > If you replace \z with (\D|\z) and substitute '\1', it might work.
> > > > (You can try it out yourself.)
>
> > > > -Rob
>
> > > > Rob Biedenharn http://agileconsultingllc.com
> > > > r...@agileconsultingllc.com
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