how to get the islamic date using autoit

[nacerbaaziz]

Hello
I have a question please
How to show Islamic date in Autoit
I mean
for example
Ramadan month
And moharam month ... etc
I hope my question is clear for you
Thanks in advance

[Melba23]

nacerbaaziz,

You should know by now that we help you get your script working - not write code from scratch. What have you tried that has not worked?

M23

[nacerbaaziz]

Hello again
my dear  
@Melba23
I do not know if there is a function in the language responsible for providing the islamic date Or another method
So I could not find an entrance to try working
But I am in dire need of this
Please help me by giving me any way to get Islamic date using autoit

[Melba23]

nacerbaaziz,

Quote

is a function in the language responsible for providing the islamic date

No.

Have you searched for an algorithm in any language which will allow you to convert normal dates into islamic dates? If so let us know and we can try and convert it into AutoIt. If not - than I suggest you do so rather then expect us to do it for you.

M23

[nacerbaaziz]

this is a function to converte the date to hijri date using cpp

 

I do not understand ith cpp language so I hope you will help me in converting this function to autoit language
 

 

void gregorian_to_hijri(int* h_y, int* h_m, int* h_d, int  g_y, int  g_m, int  g_d)
{
    int year, month, day;

    int zyr;
    int zd;
    int zm;
    int zy;

    float zjd;
    int zl;
    int zn;
    int zj;

    year = g_y;
    month = g_m;
    day = g_d;

    zyr = year;
    zd = day;
    zm = month;
    zy = zyr;

    if((zy > 1582) || ((zy == 1582) && (zm > 10)) || ((zy == 1582) && (zm == 10) && (zd > 14)))
    {
        zjd = ((1461 * (zy + 4800 + ((zm - 14) / 12))) / 4)
            + ((367 * (zm - 2 - 12 * (((zm - 14) / 12)))) / 12)
            - ((3 * (((zy + 4900 + ((zm - 14) / 12)) / 100))) / 4) + zd - 32075;
    }
    else
    {
        zjd = 367 * zy - ((7 * (zy + 5001 + ((zm - 9) / 7))) / 4)
            + ((275 * zm) / 9) + zd + 1729777;
    }

    zl = zjd - 1948440 + 10632;
    zn = ((zl - 1) / 10631);
    zl = zl - 10631 * zn + 354;
    zj = (((10985 - zl) / 5316)) * ((int)((50 * zl) / 17719))
        + ((zl / 5670)) * ((int)((43 * zl) / 15238));

    zl = zl - (((30 - zj) / 15)) * (((17719 * zj) / 50))
        - ((zj / 16)) * (((15238 * zj) / 43)) + 29;

    zm = ((24 * zl) / 709);
    zd = zl - ((709 * zm) / 24);
    zy = 30 * zn + zj - 30;

    *h_y = zy;
    *h_m = zm;
    *h_d = zd;
}

Edited June 7, 2018 by nacerbaaziz

[jchd]

There are many variants of islamic calendar. You should read up the net to choose one "standard" and try to implement it in AutoIt as I don't know of an existing published UDF. For instance the .NET frame work, Java 8 and Visual Basic have made API available for this purpose.

[JLogan3o13]

5 minutes ago, nacerbaaziz said:

this is a function to converte the date to hijri date using cpp

+

Great, that code is very translatable to AutoIt, try to do so and share your results if you get stuck.

[nacerbaaziz]

Sorry but I do not understand CPP language

[JLogan3o13]

Then I guess you have some reading to do. As Melba stated above, we are getting tired of you putting forth zero effort and demanding people write code for you, and do it now. You should be able to step through that code line by line and find an AutoIt equivalent. If you're unwilling to even try to do so, let us know so we can close the topic.

[nacerbaaziz]

First I apologize to all administrators and members
I'm asking here because this forum is very compatible with screen readers
Unlike  Lots other sites, most of which use a forums script   that is not compatible enough with the screen reader
And their explanations are mostly pictures
Unfortunately, the screen reader can not read the image
So for that the search methods I can search on the Internet are very limited
I hope you're not bothered by my primitive questions.
and  please try to help me.
I wish you a happy and good life

now I'm going back to the subject of my question, If you don't mind.

I have declared the variables in a function using autoit
But unfortunately no matter how hard I tried I could not understand the if conditional used
Can you help me to converte  only the conditional if section?
That's code what I was able to write.

 

[nacerbaaziz]

func gregorian_to_hijri($g_y = @Year, $g_m = @Mon, $g_d = @mDay, $h_y = "", $h_m = "", $h_d = "")
local $year, $month, $day
local $zyr, $zd, $zm, $zy
$zjd, $zl, $zn, $zj
$year = $g_y
$month = $g_m
$day = $g_d

endFunc

[Simpel]

2 hours ago, nacerbaaziz said:

if((zy > 1582) || ((zy == 1582) && (zm > 10)) || ((zy == 1582) && (zm == 10) && (zd > 14)))
    {
        zjd = ((1461 * (zy + 4800 + ((zm - 14) / 12))) / 4)
            + ((367 * (zm - 2 - 12 * (((zm - 14) / 12)))) / 12)
            - ((3 * (((zy + 4900 + ((zm - 14) / 12)) / 100))) / 4) + zd - 32075;
    }
    else
    {
        zjd = 367 * zy - ((7 * (zy + 5001 + ((zm - 9) / 7))) / 4)
            + ((275 * zm) / 9) + zd + 1729777;
    }

I think this could be translated as:

if ((zy > 1582) Or ((zy = 1582) And (zm > 10)) Or ((zy = 1582) And (zm = 10) And (zd > 14))) Then
        zjd = ((1461 * (zy + 4800 + ((zm - 14) / 12))) / 4) + ((367 * (zm - 2 - 12 * (((zm - 14) / 12)))) / 12) - ((3 * (((zy + 4900 + ((zm - 14) / 12)) / 100))) / 4) + zd - 32075;
    else
        zjd = 367 * zy - ((7 * (zy + 5001 + ((zm - 9) / 7))) / 4) + ((275 * zm) / 9) + zd + 1729777;
endif

This is untested but may give you a hint how to translate that part. 

Regards, Conrad

[nacerbaaziz]

this is what's i tried but it give me a errors.

func gregorian_to_hijri($G_year = @Year, $g_month = @Mon, $g_day = @MDay)
local $year = $G_year, $month = $G_month, $day = $g_day
local $zjd, $zl, $zn, $zj
local $zY, $ZM, $ZD
if (($year > 1582) Or (($year = 1582) And ($month > 10)) Or (($year = 1582) And ($month = 10) And ($day > 14))) Then
$zjd = ((1461 * ($year + 4800 + (($month - 14) / 12))) / 4) + ((367 * ($month - 2 - 12 * ((($month - 14) / 12)))) / 12) - ((3 * ((($year + 4900 + (($month - 14) / 12))/ 100))) / 4) + $day - 32075
else
$zjd = 367 * $year - ((7 * ($year + 5001 + (($month - 9) / 7))) / 4) + ((275 * $month) / 9) + $day + 1729777
endif
$zl = $zjd - 1948440 + 10632
$zn = (($zl - 1) / 10631)
$zl = $zl - 10631 * $zn + 354
$zj = (((10985 - $zl) / 5316)) * ((Int((50 * $zl) / 17719)) + (($zl / 5670)) * ((int((43 * $zl) / 15238))
$zl = $zl - (((30 - $zj) / 15)) * (((17719 * $zj) / 50)) - (($zj / 16)) * (((15238 * $zj) / 43)) + 29
$zm = ((24 * $zl) / 709)
$zd = $zl - ((709 * $zm) / 24)
$zy = 30 * $zn + $zj - 30
return $zy & " / " & $zm & " / " & $zd
endFunc

 

can any one from you check it please?

thank you

 

[Simpel]

What error message(s) do you get? Most of them I find very use- and helpful.

[Juvigy]

I fixed the errors. I dont know if the function result is correct though:

ConsoleWrite(gregorian_to_hijri()&@CRLF)
Exit
func gregorian_to_hijri($G_year = @Year, $g_month = @Mon, $g_day = @MDay)
local $year = $G_year, $month = $G_month, $day = $g_day
local $zjd, $zl, $zn, $zj
local $zY, $ZM, $ZD
if (($year > 1582) Or (($year = 1582) And ($month > 10)) Or (($year = 1582) And ($month = 10) And ($day > 14))) Then
$zjd = ((1461 * ($year + 4800 + (($month - 14) / 12))) / 4) + ((367 * ($month - 2 - 12 * ((($month - 14) / 12)))) / 12) - ((3 * ((($year + 4900 + (($month - 14) / 12))/ 100))) / 4) + $day - 32075
else
$zjd = 367 * $year - ((7 * ($year + 5001 + (($month - 9) / 7))) / 4) + ((275 * $month) / 9) + $day + 1729777
endif
$zl = $zjd - 1948440 + 10632
$zn = (($zl - 1) / 10631)
$zl = $zl - 10631 * $zn + 354
$zj = (((10985 - $zl) / 5316)) * (Int((50 * $zl) / 17719)) + ($zl / 5670) * (int((43 * $zl) / 15238))
$zl = $zl - (((30 - $zj) / 15)) * (((17719 * $zj) / 50)) - (($zj / 16)) * (((15238 * $zj) / 43)) + 29
$zm = ((24 * $zl) / 709)
$zd = $zl - ((709 * $zm) / 24)
$zy = 30 * $zn + $zj - 30
return $zy & " / " & $zm & " / " & $zd
endFunc
Exit

 

[Melba23]

nacerbaaziz,

Here is my version of that algorithm:

expandcollapse popup

#cs
    void gregorian_to_hijri(int* h_y, int* h_m, int* h_d, int  g_y, int  g_m, int  g_d)
    {
    int year, month, day;

    int zyr;
    int zd;
    int zm;
    int zy;

    float zjd;
    int zl;
    int zn;
    int zj;

    year = g_y;
    month = g_m;
    day = g_d;


    zyr = year;
    zd = day;
    zm = month;
    zy = zyr;

#ce


ConsoleWrite(_Greg2Hij() & @CRLF)

Func _Greg2Hij($sGY = @YEAR, $sGM = @MON, $sGD = @MDAY)

    $iZYr = Number($sGY)
    $iZY = Number($sGY)
    $iZM = Number($sGM)
    $iZD = Number($sGD)

    #cs

        if((zy > 1582) || ((zy == 1582) && (zm > 10)) || ((zy == 1582) && (zm == 10) && (zd > 14)))
        {
        zjd = ((1461 * (zy + 4800 + ((zm - 14) / 12))) / 4)
        + ((367 * (zm - 2 - 12 * (((zm - 14) / 12)))) / 12)
        - ((3 * (((zy + 4900 + ((zm - 14) / 12)) / 100))) / 4) + zd - 32075;
        }
        else
        {
        zjd = 367 * zy - ((7 * (zy + 5001 + ((zm - 9) / 7))) / 4)
        + ((275 * zm) / 9) + zd + 1729777;
        }
    #ce

    Local $iZJd

    If ($iZY > 1582) Or ($iZY = 1582 And $iZM > 10) Or ($iZY = 1582 And $iZM = 10 And $iZD > 14) Then

        $iZJd = Int(((1461 * ($iZY + 4800 + (($iZM - 14) / 12))) / 4) + _
                ((367 * ($iZM - 2 - 12 * ((($iZM - 14) / 12)))) / 12) - _
                ((3 * ((($iZY + 4900 + (($iZM - 14) / 12)) / 100))) / 4) + $iZD - 32075)

    Else

        $iZJd = Int((367 * $iZY) - ((7 * ($iZY + 5001 + (($iZM - 9) / 7))) / 4) + _
                ((275 * $iZM) / 9) + $iZD + 1729777)

    EndIf

    #cs
        zl = zjd - 1948440 + 10632;
        zn = ((zl - 1) / 10631);
        zl = zl - 10631 * zn + 354;
        zj = (((10985 - zl) / 5316)) * ((int)((50 * zl) / 17719))
        + ((zl / 5670)) * ((int)((43 * zl) / 15238));

        zl = zl - (((30 - zj) / 15)) * (((17719 * zj) / 50))
        - ((zj / 16)) * (((15238 * zj) / 43)) + 29;

        zm = ((24 * zl) / 709);
        zd = zl - ((709 * zm) / 24);
        zy = 30 * zn + zj - 30;

        *h_y = zy;
        *h_m = zm;
        *h_d = zd;
        }

    #ce

    Local $iZL = Int($iZJd - 1948440 + 10632)
    Local $iZN = Int(($iZL - 1) / 10631)
    $iZL -= (10631 * $iZN) + 354

    Local $iZJ = Int((((10985 - $iZL) / 5316)) * (Int((50 * $iZL) / 17719)) + (($iZL / 5670)) * (Int((43 * $iZL) / 15238)))

    $iZL -= Int((((30 - $iZJ) / 15)) * (((17719 * $iZJ) / 50)) - (($iZJ / 16)) * (((15238 * $iZJ) / 43)) + 29)

    Local $iZM =Int((24 * $iZL) / 709)
    Local $iZD = Int($iZL - ((709 * $iZM) / 24))
    Local $iZY = Int((30 * $iZN) + $iZJ - 30)

    Return $iZY & "/" & $iZM & "/" & $iZD

EndFunc   ;==>_Greg2Hij

It gives the same nonsense result as Juvigy's: 1440.79552344311 / -36.2656331431983 / 0

I think this is caused by a problem with operator precedence in those calculations. Most languages work on roughly the same pattern but in multiple operator chains like this algorithm it is easy to get the parentheses wrong.

Corrected (I think) for the Int typing - now gives: 1462/3273/26

Which is still wrong, but closer to what online convertors give as 1439/09/25

M23

Edited June 8, 2018 by Melba23
Amended variable typing

[water]

In VB it is quite easy as the following code snippet shows:

Sub test()
Dim a, b As String
a = "2017/7/7"
b = "2017/5/1"
Debug.Print DGreg("1438/12/10")
Debug.Print DHijri(b)
Debug.Print DateDiff("d", a, b)
End Sub

Function DGreg(dtHijDate As String) As Date
VBA.Calendar = vbCalHijri
DGreg = dtHijDate
VBA.Calendar = vbCalGreg
End Function

Function DHijri(dtGregDate As String) As Date
VBA.Calendar = vbCalGreg
DHijri = dtGregDate
VBA.Calendar = vbCalGreg
End Function

https://social.msdn.microsoft.com/Forums/en-US/9d36abaa-8a10-401f-bf38-fe105372049d/vba-code-for-hijri-date-converter?forum=isvvba

[jchd]

The seemingly wrong result is caused by AutoIt variables not being strongly typed.

To work as expected one need to insert Int() everytime an operation yields a partial result as double, while in C when a variable is declared int and assigned a double result, casting to int is automagic.

[Melba23]

jchd,

Thanks for that tip. I adjusted the algorithm I posted above - but obviously not correctly as it still gives a nonsense answer for the month.

M23

[nacerbaaziz]

Welcome, everyone.
First I would like to thank you for the interaction

I tried using the function that Melba23 did it.
it give me this result

1462/3273/26
But the correct date is
1439 / 09 / 23
And I didn't know why it give me a false result
Can any one of you help me to correct it?
Or if someone of you knows the VB language trying to convert the function that mister Water shared with us

Thank you already.
And I apologize to you for my stupidity.