How to get the days of Particular date?

UNIX for Advanced & Expert Users
1
2

Refer thread

3

Thanks for reply
I have seen the thread which you have mentioned
can you please explain briefly

 
d="20120103"

day=${d#??????}
temp=${d#????}
month=${temp%??}
year=${d%????}

I wont understand what is #??? refer to ?

4

These are parameter substitution that you can perform in any POSIX shell.

Here is the definition from Korn Shell manual:

${parameter#pattern}
${parameter##pattern}
                               If the shell pattern matches the beginning of
                               the value of parameter, the value of this
                               substitution is the value of the parameter
                               with the matched portion deleted; otherwise
                               the value of this parameter substituted.  In
                               the former case, the smallest matching
                               pattern is deleted; in the latter case, the
                               largest matching pattern is deleted.
${parameter%pattern}
${parameter%%pattern}
                               If the shell pattern matches the end of the
                               value of parameter, the value of parameter
                               with the matched part is deleted; otherwise
                               substitute the value of parameter.  In the
                               former, the smallest matching pattern is
                               deleted; in the latter, the largest matching
                               pattern is deleted.

So in day=${d#??????} matches 6 characters from beginning and is deleted.

I hope this helps.

5

It might be a good alternative: return 0 for Sunday, 1 for Monday etc.

echo "(6 - $( cal $MM $YY | awk 'NR==3' |wc -w ) + ( $DD % 7 ) ) %7" | bc