Date conversion in a number format

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Md
Md 2014 年 3 月 27 日
コメント済み: Walter Roberson 2014 年 3 月 27 日
Hey guys
I have an excel file with the date format like this: '03/06/2014 11:39:04.324 PM' I want to convert this date format into (yyyymmddhhmmssSSS)20140306233904324. If the time is PM, then the hour should be in the format of 24 hours. Thanks in advance.

採用された回答

Chandrasekhar
Chandrasekhar 2014 年 3 月 27 日
clc
str = '03/06/2014 11:39:04.324 PM';
[date,rem] = strtok(str,'/');
[month,rem]= strtok(rem,'/');
[year,rem]=strtok(rem,' ');
year = regexprep(year,'/','');
[hours,rem]=strtok(rem,':');
hours = regexprep(hours,' ','');
[mins,rem] = strtok(rem,':');
[sec,rem]=strtok(rem,'.');
sec = regexprep(sec,':','');
[msec,rem]= strtok(rem,' ');
[msec,y] = strtok(msec,'.');
if(strcmp(rem,' PM'))
hours = str2num(hours);
hours = hours+12;
hours = num2str(hours);
end
totaltime = [year date month hours mins sec msec]

その他の回答 (1 件)

Walter Roberson
Walter Roberson 2014 年 3 月 27 日
datestr(datevec('03/06/2014 11:39:04.324 PM', 'MM/DD/YYYY hh:mm:ss.fff PM'), 'YYYYMMDDhhmmssfff')
  2 件のコメント
Chandrasekhar
Chandrasekhar 2014 年 3 月 27 日
so simple..
Walter Roberson
Walter Roberson 2014 年 3 月 27 日
You could also use datenum instead of datevec but I think datevec should be more resistant to round-off.

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