Calculate how much of a N hour segment is outside business hours 8AM -> 9PM
I need a function that given two dates A and B which are always N hours apart (3 or 4, but anything under 10), gives me the total number of hours that fall outside of a window 8AM->5PM (or, 800 -> 1700). You can assume that the window is always small enough so that it can only start before 8AM OR end after 5PM. However, I would love to see something powerful enough to consider both cases.
I am developing an overtime calculator for an invoice, and was given this requirement.
I started playing with getHours() to see what the hour is on date A, and if it's less than 5, calculate the number of hours difference by creating a new date object with A5 = new Date(date.getTime()) and then A5.setHours(5) and then Math.abs(A - A5).
If A is more than 5, then I do the same thing but considering 8 the cutoff time.
I would need to use getHours() to check if the segment A -> B is TOTALLY outside of the 8AM->5PM window (in which case, return 0).
But... what if the window is 10 hours, 11:00PM to 9:00AM? The first IF would not happen as 11 > 5; the second IF wouldn't either, because 9 < 17.
Surely there are established, elegant ways of doing this?
Simple attempt:
function c () {
// var D = 1542695151752 // 14:31
// var D = 1542684151000 // 11:22
// var D = 1542645151000 // 00:32
var D = 1542667151000 // 6:39
var S = 8
var E = 17
var A = new Date(D)
var B = new Date(D + 1000 * 60 * 60 * 4)
var Ah = A.getHours()
var Bh = B.getHours()
var NE
var NS
if (Ah >= S && Ah <= E && Bh >= S && Bh <= E) return (Math.abs(B - A)) / 1000 / 60 / 60
if ((Ah < S || Ah > E) && (Bh < S || Bh > E)) return (0)
if (Ah >= S && Ah <= E) {
NS = A
NE = new Date(B.getTime())
NE.setHours(E)
NE.setMinutes(0)
NE.setSeconds(0)
}
if (Bh >= S && Bh <= E) {
NE = B
NS = new Date(A.getTime())
NS.setHours(S)
NS.setMinutes(0)
NS.setSeconds(0)
}
return (Math.abs(NS - NE) / 1000 / 60)
}
console.log(c())
javascript
asked Nov 20 '18 at 6:10
MercMerc
7,007114789
add a comment |
I need a function that given two dates A and B which are always N hours apart (3 or 4, but anything under 10), gives me the total number of hours that fall outside of a window 8AM->5PM (or, 800 -> 1700). You can assume that the window is always small enough so that it can only start before 8AM OR end after 5PM. However, I would love to see something powerful enough to consider both cases.
I am developing an overtime calculator for an invoice, and was given this requirement.
I started playing with getHours() to see what the hour is on date A, and if it's less than 5, calculate the number of hours difference by creating a new date object with A5 = new Date(date.getTime()) and then A5.setHours(5) and then Math.abs(A - A5).
If A is more than 5, then I do the same thing but considering 8 the cutoff time.
I would need to use getHours() to check if the segment A -> B is TOTALLY outside of the 8AM->5PM window (in which case, return 0).
But... what if the window is 10 hours, 11:00PM to 9:00AM? The first IF would not happen as 11 > 5; the second IF wouldn't either, because 9 < 17.
Surely there are established, elegant ways of doing this?
Simple attempt:
function c () {
// var D = 1542695151752 // 14:31
// var D = 1542684151000 // 11:22
// var D = 1542645151000 // 00:32
var D = 1542667151000 // 6:39
var S = 8
var E = 17
var A = new Date(D)
var B = new Date(D + 1000 * 60 * 60 * 4)
var Ah = A.getHours()
var Bh = B.getHours()
var NE
var NS
if (Ah >= S && Ah <= E && Bh >= S && Bh <= E) return (Math.abs(B - A)) / 1000 / 60 / 60
if ((Ah < S || Ah > E) && (Bh < S || Bh > E)) return (0)
if (Ah >= S && Ah <= E) {
NS = A
NE = new Date(B.getTime())
NE.setHours(E)
NE.setMinutes(0)
NE.setSeconds(0)
}
if (Bh >= S && Bh <= E) {
NE = B
NS = new Date(A.getTime())
NS.setHours(S)
NS.setMinutes(0)
NS.setSeconds(0)
}
return (Math.abs(NS - NE) / 1000 / 60)
}
console.log(c())
javascript
asked Nov 20 '18 at 6:10
MercMerc
7,007114789
This is such a basic question and yet I am going insane with it... grrrrrrr. I know there is a super-easy solution. I know it.
– Merc
Nov 20 '18 at 6:22
1
looks like a good time to use moment.js
– charlietfl
Nov 20 '18 at 6:43
Yeah, do not reinvent the wheel here. Use moment.js. Relevant xkcd, even if zones aren't your concern right now.
– cd-rum
Nov 20 '18 at 6:58
Happy to use moment! But... does it do what I need?
– Merc
Nov 20 '18 at 7:07
It's all fun games saying "Don't reinvent the wheel! Use moment!" And provide zero information about how to actually do so.
– Merc
Nov 21 '18 at 0:08
add a comment |
I need a function that given two dates A and B which are always N hours apart (3 or 4, but anything under 10), gives me the total number of hours that fall outside of a window 8AM->5PM (or, 800 -> 1700). You can assume that the window is always small enough so that it can only start before 8AM OR end after 5PM. However, I would love to see something powerful enough to consider both cases.
I am developing an overtime calculator for an invoice, and was given this requirement.
I started playing with getHours() to see what the hour is on date A, and if it's less than 5, calculate the number of hours difference by creating a new date object with A5 = new Date(date.getTime()) and then A5.setHours(5) and then Math.abs(A - A5).
If A is more than 5, then I do the same thing but considering 8 the cutoff time.
I would need to use getHours() to check if the segment A -> B is TOTALLY outside of the 8AM->5PM window (in which case, return 0).
But... what if the window is 10 hours, 11:00PM to 9:00AM? The first IF would not happen as 11 > 5; the second IF wouldn't either, because 9 < 17.
Surely there are established, elegant ways of doing this?
Simple attempt:
function c () {
// var D = 1542695151752 // 14:31
// var D = 1542684151000 // 11:22
// var D = 1542645151000 // 00:32
var D = 1542667151000 // 6:39
var S = 8
var E = 17
var A = new Date(D)
var B = new Date(D + 1000 * 60 * 60 * 4)
var Ah = A.getHours()
var Bh = B.getHours()
var NE
var NS
if (Ah >= S && Ah <= E && Bh >= S && Bh <= E) return (Math.abs(B - A)) / 1000 / 60 / 60
if ((Ah < S || Ah > E) && (Bh < S || Bh > E)) return (0)
if (Ah >= S && Ah <= E) {
NS = A
NE = new Date(B.getTime())
NE.setHours(E)
NE.setMinutes(0)
NE.setSeconds(0)
}
if (Bh >= S && Bh <= E) {
NE = B
NS = new Date(A.getTime())
NS.setHours(S)
NS.setMinutes(0)
NS.setSeconds(0)
}
return (Math.abs(NS - NE) / 1000 / 60)
}
console.log(c())
javascript
asked Nov 20 '18 at 6:10
MercMerc
7,007114789
I need a function that given two dates A and B which are always N hours apart (3 or 4, but anything under 10), gives me the total number of hours that fall outside of a window 8AM->5PM (or, 800 -> 1700). You can assume that the window is always small enough so that it can only start before 8AM OR end after 5PM. However, I would love to see something powerful enough to consider both cases.
I am developing an overtime calculator for an invoice, and was given this requirement.
I started playing with getHours() to see what the hour is on date A, and if it's less than 5, calculate the number of hours difference by creating a new date object with A5 = new Date(date.getTime()) and then A5.setHours(5) and then Math.abs(A - A5).
If A is more than 5, then I do the same thing but considering 8 the cutoff time.
I would need to use getHours() to check if the segment A -> B is TOTALLY outside of the 8AM->5PM window (in which case, return 0).
But... what if the window is 10 hours, 11:00PM to 9:00AM? The first IF would not happen as 11 > 5; the second IF wouldn't either, because 9 < 17.
Surely there are established, elegant ways of doing this?
Simple attempt:
function c () {
// var D = 1542695151752 // 14:31
// var D = 1542684151000 // 11:22
// var D = 1542645151000 // 00:32
var D = 1542667151000 // 6:39
var S = 8
var E = 17
var A = new Date(D)
var B = new Date(D + 1000 * 60 * 60 * 4)
var Ah = A.getHours()
var Bh = B.getHours()
var NE
var NS
if (Ah >= S && Ah <= E && Bh >= S && Bh <= E) return (Math.abs(B - A)) / 1000 / 60 / 60
if ((Ah < S || Ah > E) && (Bh < S || Bh > E)) return (0)
if (Ah >= S && Ah <= E) {
NS = A
NE = new Date(B.getTime())
NE.setHours(E)
NE.setMinutes(0)
NE.setSeconds(0)
}
if (Bh >= S && Bh <= E) {
NE = B
NS = new Date(A.getTime())
NS.setHours(S)
NS.setMinutes(0)
NS.setSeconds(0)
}
return (Math.abs(NS - NE) / 1000 / 60)
}
console.log(c())
javascript
javascript
asked Nov 20 '18 at 6:10
MercMerc
7,007114789
asked Nov 20 '18 at 6:10
MercMerc
7,007114789
edited Nov 20 '18 at 7:08
Merc
asked Nov 20 '18 at 6:10
MercMerc
7,007114789
asked Nov 20 '18 at 6:10
MercMerc
7,007114789
asked Nov 20 '18 at 6:10
MercMerc
7,007114789
7,007114789
This is such a basic question and yet I am going insane with it... grrrrrrr. I know there is a super-easy solution. I know it.
– Merc
Nov 20 '18 at 6:22
1
looks like a good time to use moment.js
– charlietfl
Nov 20 '18 at 6:43
Yeah, do not reinvent the wheel here. Use moment.js. Relevant xkcd, even if zones aren't your concern right now.
– cd-rum
Nov 20 '18 at 6:58
Happy to use moment! But... does it do what I need?
– Merc
Nov 20 '18 at 7:07
It's all fun games saying "Don't reinvent the wheel! Use moment!" And provide zero information about how to actually do so.
– Merc
Nov 21 '18 at 0:08
add a comment |
This is such a basic question and yet I am going insane with it... grrrrrrr. I know there is a super-easy solution. I know it.
– Merc
Nov 20 '18 at 6:22
1
looks like a good time to use moment.js
– charlietfl
Nov 20 '18 at 6:43
Yeah, do not reinvent the wheel here. Use moment.js. Relevant xkcd, even if zones aren't your concern right now.
– cd-rum
Nov 20 '18 at 6:58
Happy to use moment! But... does it do what I need?
– Merc
Nov 20 '18 at 7:07
It's all fun games saying "Don't reinvent the wheel! Use moment!" And provide zero information about how to actually do so.
– Merc
Nov 21 '18 at 0:08
This is such a basic question and yet I am going insane with it... grrrrrrr. I know there is a super-easy solution. I know it.
– Merc
Nov 20 '18 at 6:22
This is such a basic question and yet I am going insane with it... grrrrrrr. I know there is a super-easy solution. I know it.
– Merc
Nov 20 '18 at 6:22
1
1
looks like a good time to use moment.js
– charlietfl
Nov 20 '18 at 6:43
looks like a good time to use moment.js
– charlietfl
Nov 20 '18 at 6:43
Yeah, do not reinvent the wheel here. Use moment.js. Relevant xkcd, even if zones aren't your concern right now.
– cd-rum
Nov 20 '18 at 6:58
Yeah, do not reinvent the wheel here. Use moment.js. Relevant xkcd, even if zones aren't your concern right now.
– cd-rum
Nov 20 '18 at 6:58
Happy to use moment! But... does it do what I need?
– Merc
Nov 20 '18 at 7:07
Happy to use moment! But... does it do what I need?
– Merc
Nov 20 '18 at 7:07
It's all fun games saying "Don't reinvent the wheel! Use moment!" And provide zero information about how to actually do so.
– Merc
Nov 21 '18 at 0:08
It's all fun games saying "Don't reinvent the wheel! Use moment!" And provide zero information about how to actually do so.
– Merc
Nov 21 '18 at 0:08
add a comment |
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This is such a basic question and yet I am going insane with it... grrrrrrr. I know there is a super-easy solution. I know it.
– Merc
Nov 20 '18 at 6:22
1
looks like a good time to use moment.js
– charlietfl
Nov 20 '18 at 6:43
Yeah, do not reinvent the wheel here. Use moment.js. Relevant xkcd, even if zones aren't your concern right now.
– cd-rum
Nov 20 '18 at 6:58
Happy to use moment! But... does it do what I need?
– Merc
Nov 20 '18 at 7:07
It's all fun games saying "Don't reinvent the wheel! Use moment!" And provide zero information about how to actually do so.
– Merc
Nov 21 '18 at 0:08