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Thread: Math.round question

12052012, 08:14 AM #1
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Math.round question
Hello guys,
A newbie here and newbie in javascript. I'm just learning and really can't do much but read the code, understand and do some changes. Will much appreciate your help here.
So the problem I have is with rounding. Here is the little code I have (there is more and if requested I can post it here, but my question is more general):
Code:days = calculateDayCount(pickupDate, dropoffDate); getPrices(days); root_days = Math.round(days); days_count = root_days; $j('#day_count').val(days_count); $j('#header_days').html(days_count); return ["ok", 0, 0];
So what the Math.round(days) does, is it rounds as usual and stores in root_days, but the problem is that I need it rounded not as usual. Here is why:
The code will work for a service calculation and if service is provided for more then 2 hours, the service charges for the whole day. Example
1 day 1.5 hours  we'll charge for 1 day
1 day 2.5 hours  we'll charge for 2 days
2 day 0.5 hours  we'll charge for 2 days
2 days 5 hours we'll charge for 3 days
etc etc..
So the question is that how can i round appropriate way to calculate charges properly. I assume 1 hour is appox. 0.04 day.
Thank you guys.. again if you think you need the full code i can post here with the first request.
p.s. I'm thinking that i need to use math.floor here somehow, but can't really think how..

12052012, 08:51 AM #2
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In vanilla Javascript,
Code:<script type = "text/javascript"> var days = 2.03; // 2 hours = .0833333 days var d = parseInt(days) var fraction = days  d; if (fraction < .08333333) { days ; } d = Math.ceil(days); alert ("Charge is for " + d + " days"); </script>
“Everything should be as simple as it is, but not simpler.”
 both quotes Albert Einstein (German born American Physicist who developed the special and general theories of relativity. Nobel Prize for Physics in 1921. 18791955)
All the code given in this post has been tested and is intended to address the question asked.
Unless stated otherwise it is not just a demonstration.

12052012, 09:42 AM #3
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And simply ?
Code:var d = Math.floor(days+11/12);
Last edited by 007julien; 12052012 at 10:08 AM.


12052012, 09:45 AM #4
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The JavaScript function for getting the integer portion of a number is Math.floor() so that statement should read:
Code:var d = Math.floor(days);
Stephen
Learn Modern JavaScript  http://javascriptexample.net/
Helping others to solve their computer problem at http://www.felgall.com/
Don't forget to start your JavaScript code with"use strict";
which makes it easier to find errors in your code.

12052012, 01:15 PM #5
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@007julien can you please explain what exactly this Math.floor(days+11/12) achieves I'm asking this because...
@felgall & Philip M variables days_count and root_days are used elsewhere in the code also, so I thought i can do this (I know i can assign d to anyting after this, but still didn't want to confuse myself)
Code:if (days%1<0.11) {root_days = Math.floor(days)} else{ root_days = Math.floor(days)+1;} days_count = root_days;
Oh and I noticed I can use Math.ceil in place of second math.floor.

12052012, 01:24 PM #6
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11/12 of a day is 22 hours, i.e. 2 hours less than a day. If you add 11/12 to your "days" variable and then cut off the decimals, you'll end up with the correct value
Examples:
Code:days=2.03 days+11/12 = 2.9466667 Math.floor(days+11/12) = 2 (correct) days=2.084 days+11/12 = 3,0006667 Math.floor(days+11/12) = 3 (correct, because .084 days is more than 2 hours)

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mh_and (12052012)

12052012, 01:27 PM #7
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Originally Posted by mh_and;1297413days;[/CODE
var d = Math.floor(days+11/12);
11/12 of a day is (242) = 22 hours, so if you add 11/12 to the value of days (say 2.09) you get 3.00666666. Math.floor then resolves to 3 (days).
Likewise if days = 2.08 adding 11/12 gives 2.9966666, which resolves to 2 days.
That is what you asked for.
All the code given in this post has been tested and is intended to address the question asked.
Unless stated otherwise it is not just a demonstration.

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mh_and (12052012)

12052012, 03:31 PM #8
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Don't ask for help and then ignore it.
Now I see how beautifully simple is 007julien's code..
Thank you all..