Hi Simon,
Speaking of iPhones, and quite a bit OT, I saw a news piece about
photos taken with Smartphones. It appears they store metadata
including the GPS coordinates of where the picture was taken, if
that hasn't been turned off in the phone. The news reporter
checked some FB images, and found out where a young girl lived,
and the park she played at, from the images. When he went to the
house and talked with the child's mother, she was shocked,
mortified and furious, as I would have been. Needless to say, this
scares the hell out of me.
I checked some images on the net, and with an easy-to-get browser
addon, I was able to see latitude and longitude where the photos
were taken. I check some FB images, and they must have gone
through and removed the coordinates, since I couldn't find any,
and got a message that some of the metadata may have been removed.
If so, kudos to FB for doing that.
Moral of the story, apparently: don't take pictures anywhere you
don't want the world knowing you've been.
Best Regards,
Charlie
On 11-11-2010 1:33 PM, Simon Verona wrote:
Charlie
You're right - scrap my answer ! Was reading on my iPhone and
misread the OP's question.
Regards
Simon
=========================
Simon Verona
Director
Dealer Management Services Ltd
T: 0845 686 2300
Sent from my iPhone
Firstly, I think the OP's
premise is incorrect. If VAR.1 = 52.6766 and he wants to
round to 2 decimal places, the result would be 52.68, NOT
52.65. If VAR.1 = 52.6466, the result would be 52.65, not
52.60.
Second, X=INT(X*20+.5)/20 returns 52.7 (1 decimal place),
and is a little hard to interpret. Any time I do this, I
always have to think about what to multiply and divide by. I
use this form:
Round to 1, then 2 decimal places (work with the OCONV
version). If I goofed something up here, let me know.
Code:
VAR.1 = 52.6466
VAR.1.ROUNDED.1 = VAR.1 1
VAR.1.ROUNDED.2 = VAR.1 2
X = INT(VAR.1 * 20 + .5) / 20
*
CRT 'VAR.1 = ': VAR.1
CRT 'VAR.1.ROUNDED.1 = ': VAR.1.ROUNDED.1
CRT 'VAR.1.ROUNDED.2 = ': VAR.1.ROUNDED.2
CRT 'X = ': X
*
VAR.1 = 52.6766
VAR.1.ROUNDED.1 = VAR.1 1
VAR.1.ROUNDED.2 = VAR.1 2
X = INT(VAR.1 * 20 + .5) / 20
*
CRT 'VAR.1 = ': VAR.1
CRT 'VAR.1.ROUNDED.1 = ': VAR.1.ROUNDED.1
CRT 'VAR.1.ROUNDED.2 = ': VAR.1.ROUNDED.2
CRT 'X = ': X
Result:
VAR.1 = 52.6466
VAR.1.ROUNDED.1 = 52.6
VAR.1.ROUNDED.2 = 52.65
X = 52.65
VAR.1 = 52.6766
VAR.1.ROUNDED.1 = 52.7
VAR.1.ROUNDED.2 = 52.68
X = 52.7
Using the scaling format works every time, and no thinking.
Just my $.02 worth.
Charlie Noah
Charles W. Noah Associates
[email protected]
The views and opinions expressed herein are
my own (Charlie Noah) and do not necessarily reflect the
views, positions or policies of any of my former, current
or future employers, employees, clients, friends, enemies
or anyone else who might take exception to them.
On 11-11-2010 12:14 PM, Simon Verona wrote:
Try
X=INT(X*20+.5)/20
=========================
Simon Verona
Director
Dealer Management Services Ltd
T: 0845 686 2300
Sent from my iPhone
Hi,
Please help to solve the roundoff
problem. Let us take an example VAR.1 = '52.6766',
now the second character to the decimal value is 7
which is greater than 5 so the result should be get
like 52.65 and if the
second decimal point is 4 instead of 7 then the output
should be like 52.60. So
the second decimal character should be roundoff
between 0 and 5 only. I tried it by using the
substring function and i am unable to replace it to
the second character. Please help to proceed, coding
part is given below for your reference.
PROGRAM TEST.SUBSTRINGS
$INCLUDE GLOBUS.BP I_COMMON
$INCLUDE GLOBUS.BP I_EQUATE
VAR.1 = '52.6766'
VAR.2 = FIELD(VAR.1,".",2)
VAR.3 = SUBSTRINGS(VAR.2,2,1)
IF VAR.3 LT '5' THEN
VAR.3 = '0'
END ELSE
VAR.3 = '5'
END
* From here i am strucking, to proceed further
END
Thanks and Regards,
Hari Prasath. D.I
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