Date arithmetic in Unix shell scripts

Question

I need to do date arithmetic in Unix shell scripts that I use to control the execution of third party programs.

I'm using a function to increment a day and another to decrement:

IncrementaDia(){
echo $1 | awk '
BEGIN {
        diasDelMes[1] = 31
        diasDelMes[2] = 28
        diasDelMes[3] = 31
        diasDelMes[4] = 30
        diasDelMes[5] = 31
        diasDelMes[6] = 30
        diasDelMes[7] = 31
        diasDelMes[8] = 31
        diasDelMes[9] = 30
        diasDelMes[10] = 31
        diasDelMes[11] = 30
        diasDelMes[12] = 31
}
{
        anio=substr($1,1,4)
        mes=substr($1,5,2)
        dia=substr($1,7,2)

        if((anio % 4 == 0 && anio % 100 != 0) || anio % 400 == 0)
        {
                diasDelMes[2] = 29;
        }

        if( dia == diasDelMes[int(mes)] ) {
                if( int(mes) == 12 ) {
                        anio = anio + 1
                        mes = 1
                        dia = 1
                } else {
                        mes = mes + 1
                        dia = 1
                }
        } else {
                dia = dia + 1
        }
}
END {
        printf("%04d%02d%02d", anio, mes, dia)
}
'
}

if [ $# -eq 1 ]; then
        tomorrow=$1
else
        today=$(date +"%Y%m%d")
        tomorrow=$(IncrementaDia $hoy)
fi

but now I need to do more complex arithmetic.

What it's the best and more compatible way to do this?

Answer 1

Assuming you have GNU date, like so:

date --date='1 days ago' '+%a'

And similar phrases.