Date string

[Skyler]

Hi,

I am wondering if anyone could help me with a conversion. I have a date that is in the format of

string date = "19-JAN-2006";


I need to add one day to it.

Does anyone have any idea?

Thanks in advance!

[dcjr84]

I think it might be easier if didn't store the whole thing in one big string.

Something like this...

string day = "19";
string month = "January";
string year = "2006";


I say this because you are going to need to address each piece of data separately anyways. Like if the day is 31-January-2006 and you add one day to it, you will need to change it to 01-February-2006. You will also need to know how many days are in each month, account for leap years, etc, etc. :)

[Skyler]

Hi,

Thanks for the reply. Unfortunately, I have no choice in the format. It is what it is being passed to me and I have to work with it. Any ideas?

Thanks again!

[RoboTact]

Convert to any date object, add that day using correct operation on that object and print result. Don't try to add a day manually - you'll miss something.

[Skyler]

Hi,

I am sorry but I am not familiar with the date object, could you give me a sample?

Thanks Much.

[RoboTact]

I'm not familiar with that too, maybe someone here will answer better. There is standard time_t - time in seconds from some implementation-defined moment. You can convert to it from struct tm - calendar time with fields for year, day, month, etc. using make_time function, add 1 day (add 24*60*60 seconds), convert back and print. It's in C standard library. Bad advice, I'm afraid as there are days with different amount of seconds, maybe safer to add some 30 hours instead of 24.

[dcjr84]

You could do something like this. Keep in mind though that I have not done
any checking for correct number of days in month, carrying over into a new
month, leap years, etc, etc.


#include <iostream>
#include <cstring>
#include <sstream>

using namespace std;

string addDay( string date );


int main( int argc, char* argv[] ){

string date = "19-JAN-2006";
string newDate = "";

newDate = addDay( date );

cout << newDate << endl;

cin.get();

return 0;

}

//This is the function you would write
//which takes a string date as an argument
string addDay( string Date ){

string dateArray[ 3 ];
string nDate = "";
int element = 0;
int iDay, iYear;
stringstream sDay, sYear;

//This will tokenize the string into
//three parts: day, month, year
string sep = "-";
string::size_type startoftoken = Date.find_first_not_of(sep, 0);
string::size_type endoftoken = 0;

while(startoftoken != string::npos){

endoftoken = Date.find_first_of(sep, startoftoken);

if(endoftoken==string::npos)
endoftoken = Date.length();

dateArray[ element++ ] = Date.substr(startoftoken, endoftoken-startoftoken);
startoftoken = Date.find_first_not_of(sep, endoftoken);

}

//Here is where you would manipulate the date
//I.e. make sure it has the correct number of
//days, if its a leap year, etc., etc.
//dateArray[0] == day
//dateArray[1] == month
//dateArray[2] == year

//Convert day and year from string to integer
iDay = atoi( dateArray[0].c_str() );
iYear = atoi( dateArray[2].c_str() );

//Add one day
iDay++;

//Add one year, if it was 31-DEC-2006
//iYear++;

//Convert day and year from integer to string
sDay << iDay;
dateArray[0] = sDay.str();
sYear << iYear;
dateArray[2] = sYear.str();

//Concatenate new string
nDate = dateArray[0] + "-" +
dateArray[1] + "-" +
dateArray[2];

//Return our new date
return nDate;

}


This information will also help you when doing changes to your dates:

How to calculate a leap year (http://www.codetools.com/cpp/LeapYear.asp)

Number of days in each month
January - 31
February - 28 ( 29 in a leap year )
March - 31
April - 30
May - 31
June - 30
July - 31
August - 31
September - 30
October - 31
November - 30
December - 31

[Skyler]

Hi,

Thanks. This is EXACTLY what I am trying to do. Although, I am disappointed in myself for not getting it first, I hope with practice, I will get better.

~Thanks to All!