How do I remove a decimal point in value?

Drew

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Oct 7, 2003, 4:53:11 PM10/7/03

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Hi,

I'm trying to take this value 0.85 and turn it into 85, or 1.35 and
turn that into 135.

The values start off as 254.499999784 and 300, next is
#DecimalFormat(FTime/HRGoal)# and it comes out to 0.85.

One path I tried was replace, to remove the decimal but I was hoping
there was a parameter in DecimalFormat to move the decimal all the way
to right or better yet, just remove completely .. but keep the
displayed value?

Any ideas?

thnx

Charlie Griefer

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Oct 7, 2003, 5:05:43 PM10/7/03

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so you have...

and in this case, 'foo' is 0.85.

if the number will always be using the decimalFormat() function, there will
always be 2 decimal places, which means you can simply multiply the value *
100.

hth,
charlie

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Cletus Lee

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Oct 7, 2003, 5:24:24 PM10/7/03

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You can also treat it as a “.” separated list and parse and then concatenate the list members.

In article <HUFgb.64$e86....@news.uswest.net>, cha...@spamcop.com says...

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Cletus D. Lee
Bacchetta Giro
Lightning Voyager
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"Adam Haskell" <AdamH

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Oct 7, 2003, 5:54:05 PM10/7/03

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That should remove anything this is not a number including decimal points.

Adam

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Joseph Maitino

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Oct 7, 2003, 6:37:36 PM10/7/03

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sounds like the number will always be betwee 0 and 1.... so your looking
at a percentage... just multply by 100

if you just want to remove the ".' try using the replace command.

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Charlie Griefer

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Oct 8, 2003, 10:48:53 AM10/8/03

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this is probably overly picky, but I would think that parsing a list (yeah,
i know...it's just 2 elements) and concatenating what are effectively
strings, would be slower versus simply multiplying the current value * 100.

charlie

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Cletus Lee

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Oct 8, 2003, 11:29:47 AM10/8/03

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In article <qtVgb.64$JG4....@news.uswest.net>, cha...@spamcop.com says...

> this is probably overly picky, but I would think that parsing a list (yeah,
> i know...it's just 2 elements) and concatenating what are effectively
> strings, would be slower versus simply multiplying the current value * 100.
>

yes, there are many ways to solve this problem. Using regular expressions are probably the most
efficient. (wish I had thought of it)

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Cletus D. Lee

Andy J

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Oct 8, 2003, 12:24:10 PM10/8/03

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Surely the straight replace would work better in this case as you are only
looking for a ".". You dont really need to use the power of RegEx? If
anything i would expect RegEx to slow it down?

Andy J
www.andyjarrett.co.uk

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Andy J

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Oct 8, 2003, 7:33:31 PM10/8/03

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Hi, well i had to play and do a few tests. If you just use the replace() and
not reRplace its about 100% faster. Though just doing it on one string
probably wont make much difference. If anyone is interested i have put it on
my site with the results.

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Andy J
http://www.andyjarrett.co.uk

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Drew

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Oct 9, 2003, 7:59:48 AM10/9/03

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thnx for all your help guys, the replace works well