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Can I simplify this function UpdateBokningsArray by using some sort of generic formula

Started by"Tony Johansson" <johansson.andersson@telia.com>
First post2013-09-29 15:11 +0200
Last post2013-09-29 16:43 +0200
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  Can I simplify this function UpdateBokningsArray by using some sort of generic formula "Tony Johansson" <johansson.andersson@telia.com> - 2013-09-29 15:11 +0200
    Re: Can I simplify this function UpdateBokningsArray by using some sort of generic formula Ben Bacarisse <ben.usenet@bsb.me.uk> - 2013-09-29 15:30 +0100
      Re: Can I simplify this function UpdateBokningsArray by using some sort of generic formula "Tony Johansson" <johansson.andersson@telia.com> - 2013-09-29 16:43 +0200

#21461 — Can I simplify this function UpdateBokningsArray by using some sort of generic formula

From"Tony Johansson" <johansson.andersson@telia.com>
Date2013-09-29 15:11 +0200
SubjectCan I simplify this function UpdateBokningsArray by using some sort of generic formula
Message-ID<l298v8$n5$1@dont-email.me>
I have this matris that is 6 X 4 which is 6 rows and 4 columns
var bokningsArray = [["dummy", "", "", ""],
                             ["dummy", "", "", ""],
                             ["dummy", "", "", ""],
                             ["dummy", "", "", ""],
                             ["dummy", "", "", ""],
                             ["dummy", "", "", ""]];

I want to find some sort of  generic formula that I can use instead of doing 
in the way I have done in the  function UpdateBokningsArray.
Now I check if I have seatNr 1,2 or 3 I use row 0
If I have seatNr 4,5 or 6 I use row 1
If I have seatNr 7,8 or 9 I use row 2 and so on as you can see
function UpdateBokningsArray(seatNr)
        {
            switch (seatNr)
            {
                case 1:
                case 2:
                case 3:
                   bokningsArray[0][seatNr] = "BOOKED";
                   break;

                case 4:
                case 5:
                case 6:
                   bokningsArray[1][seatNr] = "BOOKED";
                   break;

                case 7:
                case 8:
                case 9:
                   bokningsArray[2][seatNr] = "BOOKED";
                   break;

                case 10:
                case 11:
                case 12:
                   bokningsArray[3][seatNr] = "BOOKED";
                   break;

                case 13:
                case 14:
                case 15:
                   bokningsArray[4][seatNr] = "BOOKED";
                   break;

                case 16:
                case 17:
                case 18:
                   bokningsArray[5][seatNr] = "BOOKED";
                   break;
            }
        }

//Tony 

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#21462

FromBen Bacarisse <ben.usenet@bsb.me.uk>
Date2013-09-29 15:30 +0100
Message-ID<0.74438714b5f725435410.20130929153005BST.87had3vgya.fsf@bsb.me.uk>
In reply to#21461
"Tony Johansson" <johansson.andersson@telia.com> writes:

> I have this matris that is 6 X 4 which is 6 rows and 4 columns
> var bokningsArray = [["dummy", "", "", ""],
>                             ["dummy", "", "", ""],
>                             ["dummy", "", "", ""],
>                             ["dummy", "", "", ""],
>                             ["dummy", "", "", ""],
>                             ["dummy", "", "", ""]];
>
> I want to find some sort of  generic formula that I can use instead of
> doing in the way I have done in the  function UpdateBokningsArray.
> Now I check if I have seatNr 1,2 or 3 I use row 0
> If I have seatNr 4,5 or 6 I use row 1
> If I have seatNr 7,8 or 9 I use row 2 and so on as you can see

You probably want Math.floor((seatNr - 1) / 3).  However...

> function UpdateBokningsArray(seatNr)
>        {
>            switch (seatNr)
>            {
<snip>
>                case 10:
>                case 11:
>                case 12:
>                   bokningsArray[3][seatNr] = "BOOKED";

Do you really want to set bokningsArray[3][12]?  I'd expect you want to
use something like (seatNr - 1) % 3 + 1 (since the zero element seems to
be unused and seats are numbered from 1.

>                   break;
<snip>
>        }

-- 
Ben.

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#21463

From"Tony Johansson" <johansson.andersson@telia.com>
Date2013-09-29 16:43 +0200
Message-ID<l29ea7$sta$1@dont-email.me>
In reply to#21462
Many thanks Ben just what I looked for.

//Tony
"Ben Bacarisse" <ben.usenet@bsb.me.uk> skrev i meddelandet 
news:0.74438714b5f725435410.20130929153005BST.87had3vgya.fsf@bsb.me.uk...
> "Tony Johansson" <johansson.andersson@telia.com> writes:
>
>> I have this matris that is 6 X 4 which is 6 rows and 4 columns
>> var bokningsArray = [["dummy", "", "", ""],
>>                             ["dummy", "", "", ""],
>>                             ["dummy", "", "", ""],
>>                             ["dummy", "", "", ""],
>>                             ["dummy", "", "", ""],
>>                             ["dummy", "", "", ""]];
>>
>> I want to find some sort of  generic formula that I can use instead of
>> doing in the way I have done in the  function UpdateBokningsArray.
>> Now I check if I have seatNr 1,2 or 3 I use row 0
>> If I have seatNr 4,5 or 6 I use row 1
>> If I have seatNr 7,8 or 9 I use row 2 and so on as you can see
>
> You probably want Math.floor((seatNr - 1) / 3).  However...
>
>> function UpdateBokningsArray(seatNr)
>>        {
>>            switch (seatNr)
>>            {
> <snip>
>>                case 10:
>>                case 11:
>>                case 12:
>>                   bokningsArray[3][seatNr] = "BOOKED";
>
> Do you really want to set bokningsArray[3][12]?  I'd expect you want to
> use something like (seatNr - 1) % 3 + 1 (since the zero element seems to
> be unused and seats are numbered from 1.
>
>>                   break;
> <snip>
>>        }
>
> -- 
> Ben. 

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