Setting a windows batch file variable to the day of the week

Question

I have a windows batch file that runs daily. Wish to log data into a file and want to rotate it (i.e. having at most the last 7 days worth of data).

Looked into the

Answer 1

Locale-dependent version: In some environments, the following will extract the day name from the date:

set dayname=%date:~0,3%

It assumes that the day name is the first part of %date%. Depending on the machine settings, though, the substring part (~0,3) would need to change.

A statement such as this would dump to a file with a three character day name:

set logfile=%date:~0,3%.log
echo some stuff > %logfile%

Locale-independent version: If you need it less dependent on the current machine's day format, another way of doing it would be to write a tiny application that prints the day of the week. Then use the output of that program from the batch file. For example, the following C application prints dayN where N=0..6.

#include 
#include 

int main( int argc, char* argv[] )
{
   time_t curtime;
   struct tm * tmval;

   time( &curtime );
   tmval = localtime( &curtime );
   // print dayN.  Or use a switch statement and print
   // the actual day name if you want
   printf( "day%d", tmval->tm_wday );
}

If the above were compiled and linked as myday.exe, then you could use it from a batch file like this:

for /f %%d in ('myday.exe') do set logfile=%%d.log
echo some stuff > %logfile%