将EXIF曝光时间计算为分数(Delphi) [英] Calculating EXIF exposure time as a fraction (Delphi)

问题描述

我试图将EXIF曝光时间显示为几分之一秒，但我没有得到预期的结果，可能是一个小错误.

I am trying to display EXIF exposure time as a fraction of seconds but I am not getting expected result, probably a small mistake.

我发现有两个例程可以进行计算，但是都带来不同的结果.

I have found two routines that do the calculation but both bring diffrent and wrong result.

我遇到的值是:"0.0806451612903226"的值是扩展类型.

The value I am having problem is: "0.0806451612903226" value is of type Extended.

DecimalToFractStr给我:结果是"16129/200000"， DecToFrac给我:结果是"5/62"

DecimalToFractStr give me: "16129/200000" as a result, DecToFrac gives me: "5/62" as a result

由于每个人都可以得到预期的结果，因此结果应为"1/12"(诀窍是什么).当我右键单击资源管理器中的图像并查看详细信息时，将获得此值.如何在我的应用程序中获取它?希望有人可以在这里帮助我.

Expected result should be "1/12" since everyone is getting it (what is the trick). When I right click image in explorer and view details I get this value. How do I get it in my application? Hope someone can help me here.

亲切问候 罗伊·科沃(Roy M Klever)

Kind Regards Roy M Klever

已解决！ 确定，经过一些测试，我弄清楚了.使用DecToFrac的结果，他们可以简单地将其与左侧的结果进行整数除法. 5 div 5 = 1/62 div 5 = 12，所以我剩下想要的结果1/12.

SOLVED! Ok after some testing I figured it out. Using the result from DecToFrac they simply integer divide it with the result from the left side. 5 div 5 = 1 / 62 div 5 = 12 so I am left with wanted result 1 / 12.

感谢您的所有输入.

亲切问候 罗伊·M·克莱沃(Roy M Klever)

Kind Regards Roy M Klever

procedure DecimalToFract(value: double; AllowedDecimals: integer; var num, den:
  integer);
var
  d, i: integer;
  ex: boolean;
begin
  d := Trunc(power(10, AllowedDecimals));
  num := Trunc(value * d);
  den := d;
  repeat
    ex := true;
    for i := 10 downto 2 do
      if ((num mod i) = 0) and ((den mod i) = 0) then
      begin
        num := num div i;
        den := den div i;
        ex := false;
      end;
  until ex;
end;

function DecimalToFractStr(value: double): string;
var
  num, den: integer;
begin
  DecimalToFract(value, 6, num, den);
  if den = 1 then
    result := inttostr(num)
  else
    result := inttostr(num) + '/' + inttostr(den);
end;

function Dec2Frac(f: Double): String;
var
  df: Double;
  lUpperPart: Integer;
  lLowerPart: Integer;
begin
  lUpperPart := 1;
  lLowerPart := 1;
  df := lUpperPart / lLowerPart;
  While (df <> f) do
  begin
    If (df < f) Then
      lUpperPart := lUpperPart + 1
    Else
    begin
      lLowerPart := lLowerPart + 1;
      lUpperPart := Trunc(f * lLowerPart);
    end;
    df := lUpperPart / lLowerPart;
  end;
  if lLowerPart = 1 then
    result := IntToStr(lUpperPart)
  else
    result := IntToStr(lUpperPart) + '/' + IntToStr(lLowerPart);
end;

推荐答案

1/12实际上不是该值的正确结果.

1/12 is not actually a correct result for that value.

也许它们可能具有某种快照"查找表，即该值与递增值列表中的值最接近1/12?

Maybe they perhaps have some kind of "snap to" lookup table i.e. this value is closest to 1/12 from a list of incremented values?

这篇关于将EXIF曝光时间计算为分数(Delphi)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！