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Date conversion in a number format

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Md
Md il 27 Mar 2014
Commentato: Walter Roberson il 27 Mar 2014
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.

Risposta accettata

Chandrasekhar
Chandrasekhar il 27 Mar 2014
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]

Più risposte (1)

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

Accedi per commentare.

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