A problem in calculating diffrence between two dates

1 回表示 (過去 30 日間)
googo
googo 2013 年 3 月 25 日
function totalDays=dateDiff(date1,date2)
y1=date1(3);
y2=date2(3);
diff1 = dateDiffSameY(date1,[31 12 y1])+1;
diff2 = dateDiffSameY([1 1 y2],date2);
sum=0;
for i=(date1(3)+1):(date2(3)-1)
if checkLeapYear(i) == 0 % function to cheak if a year is leap
sum = sum + 364;
else
sum = sum + 365;
end
end
totalDays = diff1 + diff2 + sum;
end
Well, I don't see the problem. I know that there are some situations that I forgotten but for example if I enter dateDiff([22 2 1732],[14 12 1799]) my function return 24701 When using daysact I get 24767 .
Where is my mistake? Very new in matlab, hope for your help.
  2 件のコメント
googo
googo 2013 年 3 月 25 日
編集済み: googo 2013 年 3 月 25 日
ok.. I think the problem was fixed by adding +1 to each sum in the loop
Jan
Jan 2013 年 3 月 25 日
編集済み: Jan 2013 年 3 月 25 日
Please use meaningful tags. "matlab" and "matlab code" do not help in a Matlab forum.
I cannot guess, what "daysact", "dateDiffSameY" and "checkLeapYear" is.
Do not shadow the builtin function "sum" by a variable of the same name. This leads to unexpected problems frequently.

サインインしてコメントする。

採用された回答

Jan
Jan 2013 年 3 月 25 日
The difference bertween dates can be calculated much easier:
function result = dateDiff(a, b)
result = datenum(a) - datenum(b);

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeMATLAB についてさらに検索

タグ

タグが未入力です。

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by